4    ( 


1^ 


DESCRIPTIVE  GEOMETRY. 


PREPARED  FOR  THE  USE  OF  THE  STUDENTS 
OF  THE 

MASSACHUSETTS  INSTITUTE  OF  TECHNOLOGY, 

BOSTON,  MASS. 


BOSTON: 

W.  J.  SciiOFiELD,  Printer,  105  Summer  Street. 

1886. 


Copyrighted,  1886,  by  Linus  Faunce. 


DESCRIPTIVE    GEOMETRY. 


CHAPTER  I. 

Elementaby  PmNciPLEa. 

1.  Descriptive  Geometry  is  the  art  of  representing  a  defi- 
nite body  in  space  upon  two  planes,  at  right  angles  with  each 
other,  by  lines  falling  perpendicularly  to  the  planes  from  all  the 
points  of  the  intersection  of  every  two  contiguous  sides  of  the 
body,  and  from  all  points  of  its  contour,  and  of  solving  all  graph- 
ical problems  involving  three  dimensions. 

2.  These  planes  are  called  coordinate  planes,  or  the  planes  of 
projection,  one  of  which  is  horizontal  and  the  other  vertical. 
P  and  Q,  Fig.  1,  represent  two  such  planes,  and  their  line  of  inter- 
section G  L  is  called  the  ground  line. 

3.  The  perpendicular  lines  from  a  point  in  space  to  the  planes 
of  projection  are  called  the  projecting  lines  of  that  point.  a  a,", 
a  a''.  Fig.  1,  are  the  projecting  lines  of  the  point  a. 

4.  Each  coordinate  plane  is  supposed  to  extend  indefinitely  in 
both  directions  from  the  ground  line;  hence  they  will  form  by 
their  intersection  four  dihedral  angles  (see  Fig.  1). 

The  first  angle  is  above  the  horizontal,  and  in  front  of  the  ver- 
tical plane. 

The  second  is  above  the  horizontal,  and  behind  the  vertical. 

The  third  is  below  the  horizontal,  and  behind  the  vertical. 

The  fourth  is  below  the  horizontal,  and  in  front  of  the  vertical. 

The  points  a,  6,  c,  and  rf,  Fig.  1,  are  in  the  first,  second,  third, 
and  fourth  angles  respectively. 


2065979 


4  DESCRIPTIVE  GEOMETRY. 

5.  In  order  that  the  two  projections  of  an  object  may  be  repre- 
sented on  the  same  sheet  of  paper,  the  upper  part  of  the  verti- 
cal plane  is  revolved  backward,  about  the  ground  line,  as  an  axis, 
until  it  coincides  with  the  horizontal  plane. 

This  being  done,  it  is  evident  that  all  that  portion  of  the  plane 
of  the  paper  which  is  in  front  of  or  below  the  ground  line  repre- 
sents not  only  that  part  of  the  horizontal  plane  which  is  in  front 
of  the  vertical,  but  also  that  part  of  the  vertical  plane  which  is 
below  the  horizontal,  while  the  portion  above  the  ground  line 
represents  that  part  of  the  vertical  which  is  above  the  horizontal, 
and  also  that  part  of  the  horizontal  which  is  behind  the  vertical 
plane. 

6.  All  points  in  the  first  and  second  angles  are  vertically  pro- 
jected above  the  ground  line ;  all  points  in  the  third  or  fourth 
angles  below  it. 

Points  in  the  first  and  fourth  angles  are  horizontally  projected 
in  front  of  the  ground  line ;  those  in  the  second  and  third 
behind  it. 

Thus  the  points  a,  b,  c,  and  t?,  situated  in  the  first,  second,  third, 
and  fourth  angles  respectively,  as  shown  in  Fig.  1,  are  represented 
on  the  plane  of  the  paper,  as  shown  in  Fig,  2. 

7.  Lines  situated  in  either  of  the  four  angles  are  represented 
by  their  projections  in  the  same  way  as  described  for  the  point. 

8.  The  two  projections  of  a  point  are  on  one  and  the  same 
straight  line  perpendicular  to  the  ground  line. 

9.  Two  projections  are  always  necessary  to  definitely  locate  a 
point  or  line  in  space. 

10.  The  distance  of  the  vertical  projection  of  a  point  from  the 
ground  line  is  equal  to  the  distance  of  the  point  itself,  in  space, 
from  the  horizontal  plane ;  and  the  distance  of  its  horizontal  pro- 
jection from  the  ground  line  is  equal  to  the  distance  of  the  point, 
in  space,  from  the  vertical  plane. 

11.  A  point  situated  upon  either  of  the  coordinate  planes  is  its 
own  ])rojection  on  that  plane,  and  its  other  projection  is  in  the 
ground  line. 

12.  A  right  line  is  determined  by  two  points;  hence,  any  two 
lines  drawn  at  pleasure,  exce})t  parallel  to  each  other  and  perpen- 
dicular to  the  ground  line,  will  represent  the  projections  of  a  line 
in  space. 


DESCRIPTIVE   GEOMETRY.  5 

13.  If  two  lines  are  parallel  in  space,  their  projections  upon 
either  plane  will  also  be  parallel. 

14.  If  a  right  line  is  perpendicular  to  either  coordinate  plane, 
its  projection  on  that  plane  will  be  a  point,  and  its  projection  on 
the  other  plane  will  be  perpendicular  to  the  ground  line, 

15.  If  a  line  be  parallel  to  either  plane,  its  projection  on  that 
plane  will  be  parallel  to  the  line  itself,  and  its  projection  on  the 
other  plane  will  be  parallel  to  the  ground  line. 

16.  If  a  line  is  parallel  to  both  planes,  or  to  the  ground  line, 
both  projections  will  be  parallel  to  the  ground  line. 

17.  When  a  line  is  parallel  to  a  plane,  its  projection  on  that 
plane  will  be  equal  to  the  true  length  of  the  line. 

18.  If  a  point  be  on  a  line,  its  projections  will  be  on  the  pro- 
jections of  the  line. 

19.  The  planes  which  contain  the  projecting  lines  of  all  points 
in  a  straight  line,— or,  in  other  words,  which  contain  the  straight 
line  and  are  perpendicular  to  the  planes  of  projection, —  are  called 
the  projecting  flanea  of  that  line. 

20.  If  two  lines  intersect  in  space,  their  projections  must  also 
intersect,  and  the  right  line  joining  the  points  in  which  the  pro- 
jections intersect  must  be  perpendicular  to  the  ground  line ;  for 
the  intersection  of  two  lines  must  be  a  point,  common  to  both 
lines,  whose  projections  must  be  on  the  horizontal  and  vertical 
projections  of  each  of  the  lines,  hence  at  their  intersections 
respectively. 

21.  Since  two  points  determine  a  straight  line,  it  may  be  deter- 
mined by  its  intersections  with  the  coordinate  planes.  These 
intersections  will  be  points,  and  are  called  the  traces  of  the  line, 
horizontal  or  vertical,  according  as  they  are  on  the  horizontal  or 
vertical  plane.  A  line  can  have  but  two  traces;  it  may  have  only 
one.  The  points  m  and  (/,  Fig.  3,  are  the  traces  of  the  line  g  vi, 
likewise  e  and/  of  the  line  e  f. 

22.  In  the  same  way,  a  plane  being  determined  by  two  lines,  it 
may  be  determined  by  its  intersections  with  the  coordinate  planes. 
These  intersections  will  be  lines,  and  they  are  called  the  traces  of 
the  plane.  A  plane  can  have  one  or  two  traces.  The  line  a  6, 
Fig.  3,  is  the  vertical  trace,  and  c  d  the  horizontal  trace  of  the 
plane  X. 

23.  The  traces  of  a  line  lying  in  a  plane  must  be  in  the  traces 


6  DESCEIPTIVE   GEOMETRY. 

of  that  plane  ;  for  every  line,  as  e  /  and  g  m,  Fig.  3,  lying  in  the 
plane  X,  must  intersect  the  coordinate  planes  somewhere  on  the 
lines  a  h  and  c  d  in  which  the  plane  X  intersects  the  coordinate 
planes. 

24.  The  two  traces  of  the  same  plane  must  necessarily  meet 
the  ground  line  in  the  same  point ;  for,  by  the  definition  of  traces, 
that  point  is  at  once  in  the  given  plane  and  in  each  of  the  coor- 
dinate planes ;  hence  it  is  at  the  intersection  of  the  three  planes, 
i.e.^  on  the  ground  line. 

If  the  plane  be  parallel  to  the  ground  line,  this  point  is  at  an 
infinite  distance ;  hence  the  two  traces  are  parallel  to  each  other, 
and  to  the  ground  line. 

25.  If  a  plane  be  parallel  to  either  coordinate  plane,  it  will 
have  but  one  trace,  which  will  be  on  the  other  plane  and  parallel 
to  the  ground  line. 

26.  If  a  plane  be  perpendicular  to  either  coordinate  plane,  its 
trace  on  the  other  plane  will  be  perpendicular  to  the  ground  line. 
If  it  be  perpendicular  to  both  coordinate  planes,  both  of  its  traces 
will  be  in  a  line  perpendicular  to  the  ground  line.  Such  a  plane 
is  called  a  profile  plane. 

27.  If  a  plane  pass  through  the  ground  line,  its  position  is  not 
determined. 

28.  Lines  and  planes  are  supposed  to  be  of  indefinite  length ; 
hence  the  projections  of  lines  and  the  traces  of  planes  may  always 
be  produced  indefinitely  in  either  direction. 

29.  If  a  right  line  is  perpendicular  to  a  plane,  its  projections 
will  be  respectively  perpendicular  to  the  traces  of  the  plane.  If 
a  plane  is  perpendicular  to  two  other  planes,  it  will  be  perpen- 
dicular to  their  line  of  intersection.  But  the  horizontal  project- 
ing plane  of  the  line  is  perpendicular  to  H  (Art.  19),  and  it  is 
also  perpendicular  to  the  given  plane,  since  it  contains  a  line  per- 
pendicular to  the  given  plane ;  hence  it  is  perpendicular  to  their 
line  of  intersection,  which  is  the  horizontal  trace  of  the  given 
plane.  But  the  horizontal  projection  of  the  given  line  passes 
through  this  trace,  hence  it  must  be  perpendicular  to  it. 

In  the  same  way  it  can  be  proved  that  the  vertical  projection 
of  the  line  is  perpendicular  to  the  vertical  trace  of  the  plane. 
The  converse  of  this  must  also  be  true, —  that  is,  if  a  plane  is 


DESCRIPTIVE   GEOMETRY.  7 

perpendicular  to  a  line,  the  traces  of  the  plane  must  be  respectively 
perpendicular  to  the  projections  of  the  line. 

Notation. 

30.  We  will  designate  a  point  in  space  by  a  small  letter,  and 
its  projections  by  the  same  letter  with  an  Ji  or  v  written  above : 
thus  a''  represents  the  horizontal,  and  a"  the  vertical  projection  of 
the  point  a. 

31.  A  line  in  space  will  be  designated  by  a  capital  letter,  and 
its  projections  by  the  same  letter  with  an  h  ot  v  written  above, 
the  same  as  with  the  point. 

In  speaking  of  a  line  in  space,  we  may  say  the  line  A,  or,  since 
a  line  is  determined  by  two  points,  the  line  a  h. 

32.  The  horizontal  coordinate  plane  will  be  designated  by  the 
capital  letter  H,  the  vertical  by  the  capital  letter  V,  and  any 
other  plane  in  space  by  any  capital  letter. 

The  traces  of  a  plane  will  be  designated  by  the  same  letter  as 
the  plane,  with  H  or  V  prefixed  :  thus,  H  P  denotes  the  horizon- 
tal, and  V  P  the  vertical  trace  of  the  plane  P. 

A  plane  in  space  is  determined  by  two  parallel  or  intersecting 
lines,  by  a  line  and  a  point,  or  by  three  points ;  hence,  we  may 
speak  of  a  plane  as  the  plane  P,  the  plane  (A,  B),  the  plane 
(A,  a),  or  the  plane  (a,  h,  c). 

33.  Given  and  required  lines,  if  visible,  are  represented  by  full 
lines ;  if  invisible,  by  short  dashes.  Auxiliary  lines  are  repre- 
sented by  long  and  short  dashes. 

Traces  of  planes,  if  .visible,  are  represented  by  full  lines;  if 
invisible,  by  one  long  and  two  short  dashes.  Auxiliary  planes, — 
that  is,  planes  which  are  used  simply  as  a  means  of  obtaining  the 
solution  of  a  problem,  —  are  represented  by  one  long  and  three 
short  dashes.    Construction  lines  are  represented  by  short  dashes. 

Examples. 

Ex.  1.  —  Draw  the  projections  of  a  line  in  the  third  angle  parallel  to  V, 
and  oblique  to  H. 

Ex.  2.  —  Of  a  line  in  the  second  angle  parallel  to  H,  and  oblique  to  V. 

Ex.  3.  —  Of  any  line  in  the  fourth  angle. 

Ex.  4. — Of  any  line  lying  in  V,  and  below  H. 

Ex.  5.  —  Of  a  line  lying  in  H,  and  behind  V. 


8  DESCRIPTIVE   GEOMETRY. 

Ex.   6.  —  Of  a  line  in  the  third  angle,  and  perpendicular  to  V. 

Ex.  7.  —  Of  a  line  in  the  fourth  angle,  and  perpendicular  to  H. 

Ex.  8.  —  Of  a  line  lying  in  a  profile  plane,  and  oblique  to  both  V 
,and  H. 

V    Ex.  9.  —  Of  a  line  crossing  the  first,  fourth,  and  third  angles. 
•    Ex.  10.  —  Of  a  line  crossing  the  second,  first,  and  fourth  angles. 

Ex.  11.  —  Of  a  line  crossings  the  first  and  third  angles. 

Ex.  12.  —  Of  a  line  crossing  the  second,  third,  and  fourth  angles. 

Ex.  13.  —  Of  two  intersecting  lines  lying  in  the  third  angle. 

Ex.  14.  —  Of  two  lines  lying  in  the  first  angle  which  do  not  intersect. 

Ex   15.  —  Draw  the  traces  of  a  plane  which  is  oblique  to  both  Vand  H. 

Ex.  16.  — Of  a  profile  plane. 

Ex.  17.  —  Of  a  plane  parallel  to  and  behind  V. 

Ex.  18.  — Of  a  plane  perpendicular  to  V,  and  oblique  to  H. 

Ex.  19. — Of  a  plane  perpendicular  to  H,  and  oblique  to  V. 

Ex.  20.  — Of  a  plane  pai-allel  to  G  L,  and  oblique  to  both  V  and  H. 


CHAPTER  II. 

Problems  Relating  to  the  Point,  Line,  and  Plane. 

34.  Problem  1. — To  find  the  traces  of  a  given  line. —  Let  the 
line  be  given  by  its  two  projections  A''  and  A\  Figs.  4  and  5. 
We  will  first  find  the  H  trace.  This,  being  in  H,  will  have  its  V 
projection  in  the  ground  line  (Art.  11),  and  it  must  also  be  on 
the  V  projection  of  the  line  (Art  18),  hence,  when  the  V  projec- 
tion of  the  line,  produced  if  necessary,  meets  the  ground  line,  as 
at  «"',  we  know  that  the  line  in  space  is  passing  through  H  at  some 
point,  which  may  be  either  in  front  of  or  behind  V,  and  since  the 
two  projections  of  a  point  must  be  in  a  line  perpendicular  to  G  L 
(Art  8),  and  in  the  two  projections  of  the  line  (Art.  18)  this 
point  will  be  where  a  perpendicular  to  G  L  from  a"  meets  the  H 
projection  of  the  line,  as  a'*.  Hence,  a''  is  the  H  trace  of  the 
line  A. 

If  we  put  H  in  the  place  of  V,  and  V  in  the  place  of  H,  in  the 
preceding  statement,  we  show  how  to  find  the  V  trace  of  the  line, 
—  that  is,  produce  A^  until  it  meets  G  L  at  6^ ;  erect  a  perpendicu- 
lar at  that  point,  and  l/\  the  point  where  it  intersects  A''',  will  be 
the  V  trace  of  the  line. 

35.  Problem  2.  — G-iven  the  traces  of  a  line  to  construct  its  pro- 
jections. —  Let  a*  and  h'\  Figs.  4  and  5,  be  the  traces ;  since  these 
points  lie  in  the  coordinate  planes,  their  V  and  H  projections 
respectivel}^  will  be  in  G  L  at  a"  and  V  (Art.  11) ;  joining  a"  with 
i'',  and  a''  with  6'',  we  have  A''  and  A'\  the  two  projections  required. 

36.  Problem  3. —  To  find  the  true  length  of  a  line  joining  two 
given  points  in  space.  — When  a  line  is  parallel  to  either  coordi- 
nate plane  its  projection  on  that  plane  is  equal  to  the  true  length 


10  DESCRIPTIVE   GEOMETRY. 

of  the  line  (Art.  17).  Hence,  we  have  only  to  revolve  the  line 
abont  an  axis  through  either  end  until  it  is  parallel  to  one  of  the 
planes,  and  find  its  projection  on  that  plane.  Let  A""  and  A'',  Fig. 
6,  be  the  projections  of  the  line  joining  the  points  a  and  h.  Revolve 
the  line  about  a  vertical  axis  through  the  point  a;  every  point  in 
the  line,  excej^t  the  one  in  the  axis,  moves  in  a  horizontal  circle, 
of  which  the  H  projection  is  a  circle  and  the  V  projection  a  line 
parallel  to  G  L.  When  the  H  projection  becomes  parallel  to 
G  L,  the  line  is  parallel  to  V  (Art.  15) ;  a"  and  a!"  do  not  move ; 
V  has  described  the  arc  of  a  circle  whose  radius  is  a^  6*,  and  is 
found  at  5/';  h"  moves  in  a  line  parallel  to  G  L,  and  is  found  per- 
pendicularly above  5/  at  h^'.  Therefore,  a''  b"  represents  the  true 
length  of  the  line. 

The  line  could  have  been  revolved  about  an  axis  perpendicular 
to  V,  until  it  was  parallel  to  H,  in  the  same  way. 

37.  Second  method.  —  The  true  length  of  a  line  may  also  be 
obtained  in  another  way  by  revolving  the  horizontal  projecting 
plane  of  the  line  about  its  horizontal  trace,  into  the  horizontal 
plane,  when  every  line  of  the  projecting  plane,  hence  the  line  in 
question,  will  be  shown  in  its  true  length  on  H. 

Let  the  line  be  given  as  in  Fig.  7.  The  horizontal  projecting 
lines  of  each  end  of  the  line  a  and  b  are  perpendicular  to  the  hori- 
zontal trace  a''  b''  of  the  horizontal  projecting  plane,  therefore  they 
will  be  so  after  revolution.  Hence,  draw  the  lines  «/'  a''  and  bf'  b^ 
perpendicular  to  «''5'',  and  make  them  equal  respectively  to  the 
lines  a"-' r  and  //'s,  which  represent  the  heights  of  the  points  in 
space  above  H  ;  the  line  joining  these  two  points  a/'  and  bj'  will 
be  the  true  length  required. 

This  result  could  have  been  obtained  by  revolving  the  vertical 
projecting  plane  until  it  coincides  with  V,  when  a''  a^  is  made 
equal  to  a} r  and  5'' 6/'  equal  to  b'' s,  the  true  length  being  a'' 5/'. 

38.  If  the  line  passes  through  the  plane  into  which  it  is 
revolved,  as  in  Fig.  8,  the  point  in  which  it  pierces  the  plane  being 
in  the  axis,  does  not  move,  and  the  two  ends  of  the  line  revolve 
in  opposite  directions;  that  is,  a'' af"  is  still  made  equal  to  a"r, 
and  b''b/'  equal  to  b''s;  but  they  are  lai-d  off  in  opjjosite  direc- 
tions, and  af'  bf',  the  true  length  of  the  line,  is  found  to  pass 
through  c'\  the  horizontal  trace  of  the  line. 


DESCRIPTIVE   GEOMETRY.  11 

89.  Problem  4,  — To  pass  a  plane  through  two  intersecting 
lines. — The  traces  of  a  line  lying  in  a  plane  must  be  in  the  traces 
of  the  plane  (Art.  23),  hence  we  have  only  to  find  the  traces  of 
each  line,  and  a  line  drawn  through  the  two  horizontal  traces  will 
be  the  horizontal  trace  of  the  plane,  and  one  through  the  vertical 
traces  will  be  the  vertical  trace  of  the  plane. 

Let  A  and  B,  Fig.  9,  be  the  two  lines  intersecting  in  e.  The 
traces  of  A  are  J''  and  a",  and  of  B  are  c*  and  d"  (Art.  34).  H  P 
is  drawn  through  b''  and  c\  and  V  P  through  a"  and  d".  They 
should  intersect  the  ground  line  in  the  same  point. 

40.  If  we  wished  to  pass  a  plane  through  three  points,  draw 
two  lines  through  the  points,  and  proceed  as  described  above  (see 
Fig.  10) ;  a,  6,  and  c  are  the  given  points. 

41.  Only  one  plane  can  be  passed  through  two  lines,  but  an 
infinite  number  can  be  passed  through  one  line.  Thus,  in  Fig.  11, 
the  planes  P,  Q,  R,  S,  etc.,  each  contain  the  line  B. 

42.  Problem  5. — G-iven  one  projection  of  a  line,  or  point, 
lying  in  a  plane,  to  find  the  other  projectioji.  — Let  A'',  Fig.  12,  be 
the  vertical  projection  of  a  line  lying  in  the  plane  P,  given  by  its 
traces  V  P  and  H  P.  The  traces  of  the  line  must  lie  in  the  traces 
of  the  plane  (Art.  23) ;  therefore,  the  vertical  trace  of  the  line  A 
must  be  at  «"  where  A''  intersects  V  P,  and  its  horizontal  projec- 
tion is  at  a'\  The  horizontal  trace  must  be  in  a  line  perpendicu- 
lar to  G  L  at  the  point  h"  where  A"  intersects  it,  and  on  H  P, 
hence  at  their  intersection  h''.  A^  drawn  through  a*  and  h''  will 
be  the  horizontal  projection  of  the  line. 

If  the  horizontal  projection  of  the  line  had  been  given,  the  ver- 
tical projection  would  be  found  in  the  same  way. 

43.  If  one  projection  of  a  point  were  given,  we  simply  draw 
any  line  through  the  point,  and  find  the  other  projection  of  the 
line  as  just  described. 

44.  If  the  line  is  horizontal,  its  vertical  projection  is  parallel  to 
the  ground  line,  and  its  horizontal  projection  is  parallel  to  the  hori- 
zontal trace  of  the  playie  in  tvhich  it  lies.  —  In  Fig.  13,  A"  is  the 
vertical  projection  of  such  a  line  lying  in  the  plane  P,  a"  and  a''  are 
found  as  in  Art.  42.  Since  A''  is  parallel  to  G  L,  b"  and  5^  are 
at  an  infinite  distance  from  a",  hence  A''  is  parallel  to  H  P. 


12  DESCKIPTIVE   GEOMETKY. 

45.  Problem  6. — Given  the  projections  of  a  pointy  or  line, 
lying  in  a  plane,  to  find  its  position  ivhen  the  plane  shall  have  been 
revolved  to  coincide  ivith  either  coordinate  plane.  —  Let  &'  (^,  Fig. 
12,  be  the  projections  of  a  point  in  the  phme  P.  If  we  revolve  the 
plane  about  its  horizontal  trace  until  it  coincides  with  H,  the  point 
will  move  in  a  plane  perpendicular  to  P,  and  hence  to  H  P,  and 
will  be  found  somewhere  in  ^''  <?,  and  at  the  same  distance  from 
H  P  that  it  was  in  space,  which  is  equal  to  the  hypothenuse  of  a 
right  angled  triangle,  of  which  c"  e  (the  height  of  the  point  above 
H)  is  one  side,  and  c''  d  (the  distance  of  its  horizontal  projection 
from  H  P)  is  the  other ;  hence  the  poii>t  revolves  to  c. 

In  revolving  a  line  it  is  only  necessary  to  find  the  revolved 
position  of  two  points,  and  the  position  of  the  line  is  determined. 

In  the  same  figure  we  have  found  the  revolved  position  of  one 
point  c  of  the  line  A,  but  the  point  h  being  in  H  P  does  not 
move  in  revolution ;  hence  5^  e  is  the  revolved  position  of  the 
line  A. 

46.  Proble^m  7.  — To  find  the  true  size  of  the  angle  made  hy 
tivo  intersecting  lines.  —  If  we  pass  a  plane  through  the  two  lines, 
and  then  revolve  the  plane  into  one  of  the  coordinate  planes,  the 
angle  between  the  lines  will  be  shown  in  its  true  size. 

Let  A  and  B,  Fig.  14,  be  the  given  lines  intersecting  at  a.  Pass 
a  plane  through  these  lines  by  Art.  39.  Its  horizontal  trace  is 
HP.  It  is  not  necessary  to  find  VP.  Revolve  this  plane  about 
H  P  into  H.  The  point  of  intersection  a  revolves  to  a, ;  the  points 
c^  and  V"  do  not  move,  hence  the  two  lines  will  be  found  at  c^  a, 
and  V-  a.  and  8  will  be  the  ano;le  souorht. 

47.  Problem  8. — To  find  the  true  size  and  shape  of  any  plane 
surface.  —  Let  the  surface  be  given  as  in  Fig.  15.  Pass  a  plane 
through  it  (to  do  this,  pass  a  plane  through  an}'-  two  of  its  edges, 
as  a  (^  and  d  <?),  H  P  is  its  horizontal  trace.  Revolve  this  plane 
about  H  P  into  H.  The  point  a  is  found  at  «/',  b  at  h\  c  at  c/', 
and  d  at  dl'.  Hence  aj"  5/'  c,*  dj'  is  the  true  size  and  shape  of  the 
surface  abed. 

48.  Problem  9.- — To  find  the  shortest  distance  from  a  jjoint  to 
a  line,  and  draw  the  projections  of  it  in  its  position  in  space.  —  Let 


DESCRIPTIVE   GEOMETRY.  13 

w,  Fig.  IG,  be  the  given  point,  and  A  the  given  line.  The  short- 
est distance  from  w  to  A  will  be  a  perpendicnlar  from  n  on  to  A. 
The  relative  positions  of  these  two  lines  will  only  be  shown  when 
the  plane  containing  them  has  been  revolved  into  one  of  the 
coordinate  planes.  Therefore,  pass  a  plane  through  the  line  A 
and  point  n  (to  do  this,  draw  a  line  through  n  parallel  to  A). 
H  P  is  the  horizontal  trace  (the  vertical  trace  is  not  needed). 
Revolve  this  plane  about  H  P  into  H.  The  point  n  revolves  to 
ti,,  the  point  d,  in  the  line  A,  to  d,,  «*  remains  stationary;  hence 
the  line  A  is  found  at  A,.  From  n,  drop  a  perpendicular  on  to 
A,,  and  this,  /i,  <?, ,  is  the  length  required.. 

To  tind  the  projections  of  this  in  its  correct  position,  revolve 
the  plane  back  to  its  original  position ;  c,  revolves  back  to  c^,  and 
n''  </'  is  the  horizontal  projection.  The  vertical  projection  of  e''  is 
found  at  c'',  hence  n"  c''  is  the  vertical  projection  sought. 

49.  Problem  10.  — To  find  the  angle  tvhic/i  a  given  line  makes 
tvifh  an  oblique  j^lane. — The  angle  is  that  made  by  the  line  with 
its  projection  on  the  oblique  plane.  This  projection,  together 
with  the  line  itself,  and  a  perpendicular  from  any  point  on  the 
line  to  the  plane,  form  a  right  angled  triffhgle  in  which  the  angle 
made  by  the  line  with  the  perpendicular  is  the  complement  of  the 
angle  sought. 

Let  A,  Fig.  17,  be  the  given  line,  and  P  the  given  plane. 

From  any  point  a  of  the  line  drop  a  perpendicular  B  on  to  the 
plane  (Art.  29).  Find  the  angle  which  A  and  B  make  with  each 
other  (Art.  46).  8  is  this  angle,  and  its  complement  j3  ia  the 
angle  sought. 

50.  Problem  11. —  To  find  the  line  of  intersection  of  tivo  planes. 
—  Since  the  line  of  intersection  lies  in  both  planes,  its  horizontal 
and  vertical  traces  must  lie  in  the  horizontal  and  vertical  traces 
of  both  planes  respectively,  hence  at  their  intersections. 

Let  P  and  Q,  Fig.  18,  be  the  given  planes.  V  P  and  V  Q  inter- 
sect at  a",  hence  a"  is  the  vertical  trace  of  the  line  of  intersection. 
V'  is  the  horizontal  trace,  since  it  is  the  intersection  of  H  P  and 
H  Q.  The  projections  A''  and  A''  of  the  line  are  found  from  its 
traces  by  Art.  35. 


14"  DESCRIPTIVE   GEOMETRY. 

The  preceding  explanation  applies  equally  well  to  Fig.  19, 
where  the  planes  are  assumed  in  a  different  position. 

51.  If  the  horizontal  traces  of  the  given  planes  are  parallel,  aa 
in  Fig.  20,  a''  is  the  vertical  trace ;  but,  since  H  P  and  H  Q  will 
never  intersect,  there  will  be  no  horizontal  trace,  hence  the  line 
must  be  a  horizontal  of  each  plane  (Art.  44),  and  A""  will  be 
drawn  through  a"  parallel  to  G  L,  and  A^  parallel  to  H  P  and 
HQ. 

52.  Let  the  plane  P,  Fig.  21,  be  intersected  by  the  horizontal 
plane  Q.  Since  the  line  lies  in  Q  and  R,  it  must  be  a  horizontal 
of  the  plane  P.  This  line  is  vertically  projected  in  the  vertical 
trace  of  the  plane  Q  at  A%  and  horizontally  projected  in  A''  par- 
allel to  H  P. 

63.  Let  the  planes  P  and  Q  be  situated  so  that  the  horizontal 
traces  do  not  intersect  within  the  limits  of  the  paper,  as  in 
Fig.  22. 

Pass  a  horizontal  auxiliary  plane  X  through  the  two  planes ; 
it  will  cut  a  horizontal  line  out  of  each  plane  (Art.  52),  B  out  of 
P,  and  C  out  of  Q.  Where  these  two  lines  intersect,  at  w,  will  be 
one  point  of  the  line  of  intersection,  a  will  be  another ;  hence  A 
is  the  line  of  intersection. 

If  the  planes  are  so  situated  that  neither  the  vertical  nor  the 
horizontal  traces  intersect  within  the  limits  of  the  paper.  Fig.  23, 
pass  two  horizontal  auxiliary  planes,  X  and  Y,  through  them. 
The  plane  X  gives  the  point  n  on  the  line  of  intersection,  as 
described  above,  and  the  plane  Y  the  point  m ;  hence  the  line  A, 
drawn  through  these  two  points,  is  the  line  required. 

Let  the  planes  be  so  situated  that  they  cross  the  ground  line  at 
the  same  point,  as  in  Figs.  24  and  25.  We  still  make  use  of  the 
same  principle  of  using  horizontal  auxiliary  planes.  By  this  means 
we  obtain  the  point  n.  The  planes  crossing  the  ground  line  at 
the  same  point  o,  their  line  of  intersection  must  pass  through  the 
ground  line  at  that  point;  hence  o  will  be  a  point  on  both  the 
vertical  and  horizontal  projections  of  the  line.  Therefore,  A"  and 
A""  are  the  projections  of  the  line  of  intersection. 

54.  Let  both  of  the  planes  be  parallel  to  the  ground  line,  but 
oblique  to  both  V  and  H,  as  in  Figs.  26  and  27. 

In  this  case  we  have  to  make  use  of  a  profile  plane  (Art.  26). 
This  profile  plane  cuts  out  of  each  oblique  plane  a  line  which  is 


DESCRIPTIVE    GEOMETRY.  15 

projected  in  the  traces  of  the  profile  plane.  The  intersection  of 
these  lines  will  be  one  point  in  the  line  of  intersection  sought. 
In  order  to  get  this  point,  it  is  necessary  to  revolve  the  profile 
plane  X  about  its  vertical  trace  into  V.  The  points  a"  and  J^  being 
in  the  axis,  do  not  move.  Every  other  point  in  the  plane,  as  c^  and 
d'\  move  in  horizontal  circles,  and  are  found,  after  revolution,  at 
(7,  and  d, .  The  revolved  position  of  the  lines  will,  therefore,  be 
B  and  C.  They  intersect  at  e%  In  counter  revolution  this  point 
is  found  at  e"  e^.  A''  and  A\  drawn  through  g"  and  «''  respectively, 
parallel  to  the  ground  line,  are  the  projections  of  the  line  of  inter- 
section of  the  planes  P  and  Q.  Note  that,  in  Fig.  27,  the  plane 
P  crosses  the  second  angle ;  also,  that  when  the  profile  plane  is 
revolved  into  V,  all  that  part  which  is  behind  V  rotates  in  the 
same  direction  as  that  which  is  in  front ;  hence  if  d^  rotates  to  c?,, 
c^  must  rotate  to  c,,  and  d,  and  c,  are  still  on  opposite  sides  of  the 
axis.  In  counter  revolution  e\  which  is  on  the  same  side  of  the 
axis  as  <?, ,  will  revolve  to  e\  and  the  line  of  intersection  is  in 
the  second  angle. 

55.  Let  one  of  the  planes  pass  through  the  ground  line. 
Both  of  its  traces  will  be  in  the  ground  line,  and  it  will  be 

necessary,  in  order  to  determine  the  plane,  to  give  the  angle  it 
makes  with  V  or  H. 

Let  P  be  given,  as  shown  in  Figs.  28  and  29,  and  Q  passing 
through  the  ground  line  and  making  angle  5  with  H.  Pass  the  pro- 
file plane  X  through  them.  It  will  cut  from  these  planes  lines  which 
are  shown  in  their  revolved  positions  at  B  and  C  respectively, 
intersecting  at  e%  In  counter  revolution  this  point  goes  to  e"  e^. 
Joining  €"  and  e''  with  o,  we  get  A""  and  A*"  as  the  two  projections 
of  the  line  of  intersection.  Note  particularly  the  construction  in 
Fig.  29. 

56.  Problem  12.  — To  find  tvhere  a  line  pierces  a  plane. —  Pass 
any  plane  through  the  line,  and  find  its  intersection  with  the 
given  plane.  Since  this  line  and  the  given  line  are  in  the  same 
plane  they  will  intersect,  unless  they  are  parallel,  and  the  point 
where  they  intersect  must  be  in  the  given  plane,  since  it  is  on  a 
line  of  that  plane  ;  hence  it  is  the  point  where  the  line  pierces 
the  plane. 

Let  A,  Fig.  30,  be  the  given  line,  and  P  the  given  plane.     We 


16  DESCRIPTIVE   GEOMETRY. 

will  take  the  plane  which  projects  A  upon  H  as  the  auxiliary 
plane.  Its  traces  are  V  X  and  H  X.  a  6  is  the  line  of  intersec- 
tion of  the  planes  X  and  P.  The  line  A  intersects  the  line  a  5  at 
the  point  c,  which  is,  therefore,  the  point  where  A  pierces  P. 

Fig.  31  shows  the  construction,  if  the  vertical  projecting  plane 
were  used  as  the  auxiliary  plane. 

57.  In  Fig.  32  the  general  method  is  shown.  Pass  any  plane 
X  through  the  line  A  (Art.  41).  Its  intersection  with  P  is  the 
line  d  e.  c,  the  intersection  of  the  lines  A  and  d  e,  is  the  point 
required. 

58.  Where  the  plane  is  determined  by  two  intersecting  lines 
the  principle  remains  the  same,  but  the  details  of  the  construction 
are  different. 

Let  A,  Fig.  33,  be  the  given  line,  and  let  the  plane  be  deter- 
mined by  the  two  lines  B  and  C  intersecting  in  the  point  a.  First 
find  the  line  of  intersection  of  the  plane  projecting  A  upon  H 
with  the  plane  (B,  C).  The  line  C  pierces  the  plane  X  at  a  point 
whose  horizontal  projection  is  c''  (where  C^  intersects  H  X).  The 
vertical  projection  of  this  point  must  be  in  the  vertical  projection 
of  the  line,  and  perpendicularly  above  or  below  it,  hence  at  c''.  In 
the  same  way  we  find  that  B  pierces  X  at  the  point  b.  Tliere- 
fore,  b"  c",  b'' c''  are  the  projections  of  this  line  of  intersection.  A 
intersects  this  line  5  c  at  the  point  d,  which  is,  therefore,  the  point 
where  it  pierces  the  plane  (B,  C). 

59.  Problem  13.  — To  find  the  shortest  distance  from  a  point  to 
a  plane.  —  The  shortest  distance  must  evidently  be  measured 
along  a  line  from  the  point  perpendicular  to  the  plane. 

Let  a,  Fig.  34,  be  the  given  point,  and  P  the  given  plane. 
From  the  point  a  drop  the  perpendicular  B  upon  P  (Art.  29). 
B  pierces  P  at  d  (Art.  56^.  B, ,  the  true  length  of  ad,  is  the  dis- 
tance required. 

60.  Since  the  definition  of  the  projection  of  a  point  on  a  plane 
is  where  a  perpendicular  through  the  point  pierces  the  plane,  the 
point  d  is  the  projection  of  the  point  a  upon  the  plane  P. 

Hence,  to  find  the  projection  of  a  line  upon  an'  oblique  plane, 
it  is  only  necessary  to  find  where  a  perpendicular  from  each  end 
of  the  line  pierces  the  plane,  and  join  the  points  so  found. 


DESCRIPTIVE  GEOMETRY.  17 

Let  A  be  the  given  line,  Fig.  35,  and  P  the  given  plane.  The 
l)rojection  of  a  upon  P  is  g,  and  of  h  upon  P  is  o,  hence  B  is  the 
projection  of  A  upon  P. 

61.  Problem  14.  —  To  pass  a  plane  through  a  given  point  par- 
allel to  a  given  plane. —  Since  the  required  plane  is  to  be  parallel  to 
the  given  plane,  their  traces  must  be  parallel ;  therefore,  the  hori- 
zontal projection  of  a  horizontal  line  of  the  required  plane  will  be 
parallel  to  the  horizontal  trace  of  the  given  plane. 

Let  a.  Fig.  36,  be  the  given  point,  and  P  the  given  plane. 
Draw  the  horizontal  line  X  of  the  required  plane  through  a.  X'' 
and  X''  are  its  projections,  X''  being  parallel  to  H  P.  If  is  the 
vertical  trace  of  this  line.  Hence  V  Q,  drawn  through  5"  parallel 
to  V  P,  and  H  Q  parallel  to  H  P,  are  the  traces  of  the  required 
plane. 

62.  Problem  15. — To  pass  a  plane  through  a  given  point  per- 
pendicular to  a  given  line.  —  Let  a,  Fig.  37,  be  the  given  point, 
and  A  the  given  line.  The  traces  of  the  plane  must  be  perpen- 
dicular to  the  projections  of  the  line  (Art.  29).  Draw  the  hori- 
zontal line  X  of  the  required  plane  through  a.  X''  and  X*"  are  its 
projections,  X''  being  perpendicular  to  A*",  b"  is  the  vertical  trace 
of  this  line.  Hence  V  P,  drawn  through  b"  perpendicular  to  A'', 
and  H  P,  perpendicular  to  A*",  are  the  traces  of  the  required  plane. 

68.  Problem  16.  — To  pass  a  plane  through  a  given  line  parallel 
to  another  given  line.  —  Let  A  and  B,  Fig.  38,  b,e  the  given  lines. 
Through  any  point  d  of  the  line  A  draw  a  line  parallel  to  B.  B/ 
and  B,*^  parallel  respectively  to  B^  and  B"^  (Art.  13)  will  be  the 
projections  of  such  a  line.  The  plane  P,  which  contains  these 
lines,  A  and  B,  (Art.  39),  will  contain  A  and  be  parallel  to  B, 
since  it  contains  a  line  which  is  parallel  to  B. 

64.  Problem  17.  — To  pass  a  plane  through  a  given  line  p>erpen- 
dicular  to  a  given  plane.  —  Let  A,  Fig.  39,  be  the  given  line,  and 
P  the  given  plane.  Through  any  point  e  of  the  line  A  draw  a 
line  perpendicular  to  P.  B"  and  B""  will  be  its  projections  (Art. 
29).  The  plane  Q  which  contains  these  lines,  A  and  B  (Art.  39), 
will  contain  A  and  be  perpendicular  to  the  plane  P,  since  it  con- 
tains a  line  which  is  perpendicular  to  P. 


18  DESCRIPTIVE  GEOMETRY. 

65.  Problem  18.  To  construct  the  projections  of  the  shortest  line 
which  can  he  drawn  terminating  in  two  right  lines  not  in  the  same 
plane.  —  Let  A  and  B,  Fig.  40,  be  the  given  lines.  The  required 
line  must  be  perpendicular  to  both  given  lines.  Pass  a  plane 
through  A  parallel  to  B  (Art.  63).  V  P  and  H  P  will  be  its  traces. 
Find  the  projection  of  B  upon  this  plane  (Art.  60).  It  being  paral- 
lel to  the  plane,  its  projection  on  the  plane  will  be  parallel  to  itself. 
Hence,  it  will  only  be  necessary  to  find  the  projection  of  one 
point,  as  m  which  is  at  r,  and  D  is  the  projection  of  B  upon  P. 
From  s,  the  intersection  of  D  and  A,  draw  a  perpendicular  E  to 
the  plane  P.  It  will  intersect  B  at  the  point  ^,  and  be  perpen- 
dicular to  both  A  and  B.     E,  will  be  its  true  length  (Art.  37). 

66.  The  line  of  greatest  declivity  of  a  plane  is  that  line  which 
makes  the  greatest  angle  with  H. 

This  is  the  line  of  intersection  a  b,  (Fig.  41),  of  the  plane  P 
with  the  plane  X,  which  is  perpendicular  to  both  P  and  H. 

Let  B  be  the  angle  a  h  makes  with  H.    Then  tan. /3=  ^^ But 

a^b^,  perpendicular  to  H  P,  is  the  shortest  line  that  can  be  drawn 
from  a''  to  H  P ;  hence,  ^  is  the  greatest  angle  that  can  be  made 
by  any  line  of  the  plane  P  with  H. 

67.  Problem  19. — To  find  the  angle  a  given  plane  makes  with 
either  plaiie  of  projection. — We  will  first  find  the  angle  made  with 

.  H.  This  angle  is  the  one  made  by  its  line  of  greatest  declivity 
with  H. 

Let  P,  Fig.  42,  be  the  given  plane.  Revolve  the  plane  X,  which 
cuts  from  P  the  line  of  greatest  declivity  (Art.  Q%^^  about  its  ver- 
tical trace  into  V.  d^  being  in  the  axis  does  not  move  ;  h''  moves 
in  the  arc  of  a  circle  with  a''  as  a  centre,  and  is  found  at  5,  and  |3 
is  the  required  angle. 

The  plane  X  could,  just  as  well,  have  been  revolved  about  its 
horizontal  trace  into  H.  In  that  case  ^'',  Fig.  42,  would  remain  sta- 
tionary, a''  would  revolve  to  a  in  a  line  perpendicular  to  a''  5*,  and 
8  would  be  the  angle  required.    Of  course,  §  and  8  should  be  equal. 

If  the  auxiliary  plane  X  were  taken  perpendicular  to  V  and  P, 
it  would  cut  from  P  the  line  making  the  greatest  angle  with  V, 
which,  being  revolved  into  one  of  the  planes  of  projection,  would 
shoAV  the  angle  the  plane  makes  with  V, 


DESCRIPTIVE   GEOMETRY.  19 

In  Fig.  43,  (3  is  the  angle  P  makes  with  V,  and  d  the  angle  it 
makes  with  H. 

68.  If  one  trace  of  a  plane  and  the  angle  it  makes  with  the 
other  coordinate  plane  were  given,  we  would  determine  the  other 
trace  by  the  reverse  of  the  last  problem.  That  is,  if  H  F  and  S, 
Fig.  42,  were  given,  to  find  VP;  draw  any  line,  as  a'' b''  perpen- 
dicular to  H  P  ;  draw  a  1/  making  angle  8  with  a''  6\  and  a  a''  jier- 
pendicular  to  a*  5* ;  rotate  this  triangle  about  H  X  as  an  axis 
until  it  is  perpendicular  to  H ;  a  is  found  at  a",  and  must  be  one 
point  of  V  P ;  join  a"  and  the  point  where  H  P  intersects  the 
ground  line,  and  VP  is  obtained. 

69.  Problem  20. — Q-iven  the  angle  a  plane  makes  loith  each 
coordinate  plane  to  construct  its  traces. — A  plane  is  tangent  to  a 
sphere  at  a  single  point.  If  we  pass  an  auxiliary  plane  through 
this  point,  the  centre  of  the  sphere,  and  perpendicular  to  H,  it 
will  cut  a  great  circle  from  the  sphere  and  a  line  from  the  tangent 
plane,  which  will  be  tangent  to  the  circle.  But  this  line  is  the 
line  of  greatest  declivity  of  the  tangent  plane,  since  it  is  a  line 
ci(it  from  it  by  a  plane  perpendicular  to  H  and  to  the  tangent  plane 
as  it  contains  a  normal  to  the  tangent  plane.  If  this  auxiliary 
plane  be  revolved  about  its  vertical  trace  to  coincide  with  V,  the 
line  and  circle  will  be  shown  in  their  true  size  and  position,  and 
also  the  true   size  of  the  angle  the  plane  makes  with  H. 

An  auxiliary  plane  passed  through  the  tangent  point,  the  centre 
of  the  sphere,  and  perpendicular  to  V,  cuts  from  the  sphere  a  great 
circle  and  from  the  tangent  plane  a  line  tangent  to  the  circle,  and 
which  makes  the  greatest  angle  with  V,  and  this  being  revolved 
into  H  shows  the  size  of  the  angle  the  plane  makes  with  V,  etc. 

In  Fig.  44,  with  o  as  a  centre,  and  any  radius  describe  a  circle. 
This  will  represent  the  revolved  position  of  both  sections  of  the 
sphere.  Draw  a  b  tangent  to  this  circle  so  that  the  angle  ab  o  is 
equal  to  /3,  the  angle  the  plane  makes  with  H.  This  line  is  the 
revolved  position  of  the  line  of  greatest  declivity  of  the  plane,  and 
a  will  be  one  point  on  the  vertical  trace  of  this  plane.  Draw  d  c 
tangent  to  the  circle  making  the  angle  c  d  o  equal  to  d,  the  angle 
which  the  plane  makes  with  V.  This  line  is  the  revolved  position 
of  the  line  of  the  plane  which  makes  the  greatest  angle  with  V, 
and  c  will  be  one  point  on  the  horizontal  trace  of  this  plane. 


20  DESCRIPTIVE   GEOMETRY. 

In  the  counter  revolution  of  a  h,  a  remains  stationary  and  b 
moves  in  the  arc  of  a  circle  of  which  o  is  the  centre  until  ab  is 
perpendicular  to  the  horizontal  trace  of  the  plane.  Therefore, 
with  0  as  a  centre,  and  o  5  as  a  radius,  describe  a  circle,  and 
through  c  tangent  to  this  circle  draw  H  P,  which  will  be  the  hori- 
zontal trace  required.  The  line  VP,  joining  a  and  the  point  where 
HP  crosses  the  ground  line,  is  the  vertical  trace  of  the  same 
plane.  VP  could  be  constructed  independently  by  drawing  it 
through  a  tangent  to  the  circle  drawn  with  o  as  a  centre,  and  o  d, 
as  a  radius. 

In  every  case  the  sum  of  the  angles  made  by  a  plane  with  each 
plane  of  projection  must  be  equal  to,  or  greater  than,  90  degrees. 

70.  Problem  21. —  To  find  the  angle  made  by  two  oblique  planes. 
—  From  any  point  in  space  let  fall  a  perpendicular  to  each  plane. 
The  supplement  of  the  angle  made  by  these  two  lines  will  be  the 
angle  required. 

71.  Problem  22. —  To  draw  the  projections  of  any  solids  of  defi' 
nite  size,  occupying  a  fixed  position  in  space,  resting  unth  its  base  on 
an  oblique  plane  which  makes  known  angles  with  both  coordinate 
planes,  and  construct  the  projections  of  its  shadow  upon  this  oblique 
plane.  —  This  problem  is  merely  an  application  of  several  of  the 
preceding  problems. 

In  Fig.  45,  let  p'  be  the  angle  the  oblique  plane  makes  with  H, 
and  8  the  angle  it  makes  with  V.  Find  V  P  and  H  P  by  Art.  69. 
Let  the  solid  be  a  rectangular  prism,  the  lowest  point  of  which  is 
at  a  distance  equal  to  a'' r  above  H,  and  a''  r  in  front  of  V.  The 
intersections  a''  and  a''  of  the  two  projections  respectively  of  the 
horizontal  line  X  of  the  plane  P,  which  is  at  a  distance  above  H 
equal  to  a"  r,  and  the  line  R  of  the  plane  P,  which  is  at  a  distance 
in  front  of  V  equal  to  t/r,  are  the  projections  of  this  lowest  point. 
Having  found  the  projections  of  one  point  of  the  base,  it  is  neces- 
sary to  revolve  the  plane  P  into  one  of  the  coordinate  planes  in 
order  to  show  the  base  of  the  prism  in  its  true  size  and  position. 
The  point  a  revolves  to  a,  (Art.  45).  Make  a,  b,  c,  d,  equal  to 
the  true  size  of  the  base  of  the  prism,  and  in  the  desired  position. 
In  counter  revolution  these  points  will  be  found  in  horizontal 
projection  at  a*,  b'',  c'',  and  d''  respectively.    The  vertical  projections 


DESCRIPTIVE   GEOMETKY.  21 

of  these  points  are  found  at  rt",  5",  c",  and  d"  (Art.  42).  This  being 
a  right  prism,  the  elements  are  perpendicular  to  the  plane,  hence 
their  projections  are  perpendicular  respectively  to  H  P  and  V  P 
(Art.  29).  To  find  n^^  the  horizontal  projection  of  the  top  end  of 
the  element  an,  revolve  the  plane  projecting  this  element  upon 
H,  with  its  line  of  intersection  with  P,  into  H.  The  line  of  inter- 
section revolves  to  V't,  the  point  a  to  s,  the  element  an  to  .s-?«, , 
perpendicular  to  V"  t,  sn,  being  equal  to  the  real  height  of  the 
prism.  In  counter  revolution  n,  goes  to  n''.  Making  the  other  ele- 
ments of  the  same  length  as  a''n'\  and  joining  the  tops,  the  hori- 
zontal projection  of  the  prism  is  completed  ;  e",  m",  n",  and  o"  are  on 
the  vertical  projections  of  the  elements  vertically  above  e'\  ??t\  ?i\ 
and  o''. 

72.  Since  the  definition  of  the  shadow  of  a  point  upon  a  plane 
is  where  a  ray  of  light  through  that  point  pierces  the  plane,  to 
find  the  shadow  of  this  prism  upon  P,  it  is  only  necessary  to  find 
where  the  ray  of  light,  as  L,  through  the  point  e  pierces  P  (Art. 
56)  at  e,;  and  do  this  for  the  other  points  which  cast  shadows. 


CHAPTER   III. 

Principles   akd   Problems   Relating   to   the   Cylinder, 
Cone,  and  Double  Curved  Surface  of  Revolution. 

73.  A  cylinder  may  be  generated  by  a  straight  line,  called  the 
generatrix^  moving  along  a  curved  line,  called  the  directrix.,  with 
all  its  positions  parallel.  The  different  positions  of  the  generatrix 
are  called  elements.  If  the  curve  has  a  centre,  a  line  drawn  through 
it,  and  parallel  to  the  elements,  is  called  the  axis  of  the  cylinder. 

Any  curved  section  of  a  cylinder  made  by  a  cutting  plane,  and 
taken  as  the  limit  of  the  elements,  may  be  called  the  base  of  the 
cylinder.  If  the  plane  is  perpendicular  to  the  axis,  the  section  is 
a  right  section^  and,  if  this  section  be  taken  as  the  base,  the  cylin- 
der is  called  a  right  cylinder.  If  the  right  section  is  a  circle,  the 
cylinder  is  a  cylinder  of  revolution. 

A  cylinder  may  be  called  circular  or  elliptic,  according  as  its 
right  section  is  a  circle  or  an  ellipse. 

74.  The  cone  differs  from  the  cylinder  only  in  the  fact  that  its 
elements,  instead  of  being  parallel,  pass  through  a  common  point 
called  the  vertex. 

The  statements  and  definitions  relating  to  the  cylinder,  in  the 
preceding  article,  are  equally  applicable  to  the  cone. 

75.  The  cylinder  and  cone  must  be  represented  in  projection 
by  their  lines  of  apparent  contour,  whatever  other  lines  may  be 
drawn  on  their  surfaces. 

These  surfaces  are  called  single  curved  surfaces. 

76.  A  double  curved  surface  of  revolution,  such  as  a  sphere, 
ellipsoid,  torus,  etc.,  is  generated  by  a  plane  curve  revolving  about 
a  right  line  contained  in  its  plane.  This  line  is  called  the  axis  of 
the  surface,  and  the  generating  curve  is  called  the  meridian  line., 
and  its  plane  the  meridian  plane.  When  a  meridian  plane  is 
parallel  to  V,  it  is  called  the  principal  meridian  plane. 


DESCRIPTIVE  GEOMETRY.  '  23 

Every  point  in  the  meridian  line  describes  the  arc  of  a  circle, 
the  centre  of  which  is  in  the  axis  of  the  surface.  This  circle  is 
called  a  parallel.  Hence  any  plane  perpendicular  to  the  axis  of 
a  surface  of  revolution  will  cut  a  parallel  from  the  surface. 

77.  The  simplest  way  of  representing  a  double  curved  surface 
of  revolution  is  to  assume  the  axis  perpendicular  to  H,  when  the 
greatest  parallel  will  be  its  horizontal  projection,  and  the  princi- 
pal meridian  will  be  its  vertical  projection. 

78.  A  plane  which  contains  two  lines  that  are  tangent  to  a 
surface  at  a  common  point  will  be  tangent  to  the  surface,  and 
will,  moreover,  contain  every  line  which  is  tangent  to  the  surface 
at  that  point. 

In  the  case  of  a  cylinder  and  cone,  the  tangent  plane  must  con- 
tain an  element  of  the  surface. 

79.  If  a  single  curved  surface  and  its  tangent  plane  be  inter- 
sected by  any  secant  plane,  the  line  cut  from  the  tangent  plane 
will  be  tangent  to  the  curve  cut  from  the  surface.  Hence  if  this 
secant  plane  happens  to  be  the  horizontal  coordinate  plane,  the 
horizontal  trace  of  the  tangent  plane  must  be  tangent  to  the 
base  of  the  surface  at  the  point  where  the  element  of  contact 
pierces  H. 

80.  A  normal  to  a  surface  at  any  point  is  a  right  line  perpen- 
dicular to  the  tangent  plane  at  that  point. 

A  normal  plane  is  a  plane  which  contains  the  normal  line,  hence 
it  will  be  perpendicular  to  the  tangent  plane. 

81.  A  plane  which  contains  one  line  which  is  tangent  to  a  sur- 
face, and  is  perpendicular  to  the  normal  at  that  point,  must  be 
tangent  to  the  surface  at  that  point. 

82.  Problem  23. — Given  one  projection  of  a  point  on  the  sur- 
face of  a  cylinder^  to  find  its  other  projection,  and,  second,  to  pass  a 
plane  tangent  to  the  cylinder  through  this  point. — Let  the  cylinder  be 
given  as  in  Fig.  46,  and  let  a''  be  the  horizontal  projection  of  a 
point  on  its  surface. 

Through  a''  draw  E*"  parallel  to  A^.  This  will  be  the  horizon- 
tal projection  of  two  elements,  one-  on  the  upper  and  one  on  the 
lower  side  of  the  cylinder.  The  base  of  the  cylinder,  since  it  rests 
on  H,  is  the  locus  of  the  horizontal  traces  of  all  the  elements  ; 
hence  d''  and  e'',  where  E''  intersects  the  circle  of  the  base,  are  the 


24  DESCRIPTIVE   GEOMETRY. 

horizontal  traces  of  the  two  elements,  and  d"'  and  e"  are  their 
vertical  projections.  E''  and  E/  drawn  through  <:7"  and  e"  respec- 
tively parallel  to  C%  are  the  vertical  projections  of  the  two  ele- 
ments horizontally  projected  in  E*".  a''  and  a,",  where  a  line 
through  a^  perpendicular  to  GL  intersects  E^  and  E/,  are  the 
vertical  projections  of  the  two  points  horizontally  projected  in  a!". 

Second.  To  drmv  a  plane  tangent  to  the  cylinder  at  the  point 
a"  a!".  —  HP,  drawn  tangent  to  the  base  of  the  cylinder  at  t?'',  will 
be  the  horizontal  trace  of  the  tangent  plane  (Art.  79).  The  line 
X  will  be  a  horizontal  of  the  plane  P  ;  its  vertical  trace  is  m" ; 
V  P,  drawn  through  nf  and  the  point  where  H  P  intersects  the 
ground  line,  is  the  vertical  trace  of  the  tangent  plane.  The  verti- 
cal trace  of  the  plane  must,  of  course,  pass  through  the  vertical 
trace  of  the  element  of  contact,  and  the  vertical  trace  of  the  tan- 
gent plane  could  have  been  found  in  that  way.  In  finding  the 
plane  Q  tangent  to  the  cylinder  at  the  point  a,"  a!\  since  H  Q  does 
not  intersect  G  L  within  the  limits  of  the  paper,  it  was  necessary 
to  use  both  methods  to  find  V  Q. 

83.  Fig.  47  shows  the  construction  when  the  axis  of  the  cylin- 
der is  parallel  to  both  V  and  H. 

Let  a!'  be  the  horizontal  projection  of  a  point  on  its  surface.  It 
is  here  necessary  to  make  use  of  a  profile  plane.  To  avoid  con- 
fusing the  figure,  assume  the  profile  plane  X,  containing  one  end 
of  the  C3dinder,  and  revolve  it  away  from  the  figure  about  either 
trace  into  one  of  the  coordinate  planes ;  in  this  case  about  V  X 
into  V.  It  cuts  out  of  the  cylinder  a  right  section  —  a  circle  in 
this  case  —  of  which  c  is  the  centre ;  c  rotates  to  c,%  and  the  cir- 
cular section  with  it.  The  horizontal  projection  of  the  element 
E''  througli  tlie  point  a  meets  the  profile  plane  at  a  point  whose 
horizontal  projection  is  V\  In  revolution  V'  moves  to  i,/';  erect- 
ing a  perpendicular  at  5,/',  it  cuts  the  circle  at  ^>„''  and  i,,/'.  These 
are  the  revolved  positions  of  the  points  in  which  the  two  possible 
elements  through  a*  meet  the  plane  X.  In  counter  revolution 
they  are  found  at  ?>''  and  J/'  respectively.  Through  these  points 
draw  the  elements  E"  and  E/.  These  will  be  the  vertical  projec- 
tions of  two  elements  horizontally  projected  at  E'';  aP  and  a/, 
where  a  line  through  a*,  perpendicular  to  G  L,  intersects  E''  and 
E/,  are  the  vertical  projections  of  the  two  points  horizontally  pro- 
jected at  a''. 


DESCRIPTIVE   GEOMETRY.  25 

To  draw  a  plane  tangent  to  the  eylinder  at  the  point  a"  a^.  —  The 
profile  plane  cuts  out  of  the  plane,  tangent  to  the  cylinder  along 
the  element  E''  E'',  a  line  which  is  tangent  to  the  curve  cut  from 
the  cylinder  at  the  point  If  h'^  where  the  element  E  pierces  X 
(Art.  79).  Hence  T,  tangent  to  the  circle  at  6„",  is  the  revolved 
position  of  this  line  of  intersection.  In  counter  revolution  d"  does 
not  move,  and  ej^  revolves  to  e'' ;  d"  and  e^  are  the  traces  of  this 
line  which  lies  in  the  tangent  plane  ;  and,  since  the  tangent  plane 
contains  the  element  E  which  is  parallel  to  both  V  and  H,  V  P 
and  H  F,  drawn  through  d"  and  e*  parallel  to  G  L,  are  the  traces  of 
the  tangent  plane. 

In  the  same  manner  the  tangent  plane  through  the  point  a/'  a* 
may  be  drawn.  Note  here  that,  in  counter  revolution,  m!"  re- 
volves to  w\  and  V  Q  and  H  Q  are  both  found  above  the  ground 
line,  which  indicates  that  the  plane  crosses  from  the  first,  through 
the  second,  and  into  the  third  angle. 

84.  Problem  24. — Given  07ie  projection  of  a  point  on  the  sur- 
face of  a  cone  to  find  its  other  projection;  and.^  second^  to  pass  a 
plane  tangent  to  the  cone  through  this  point. —  Let  the  cone  be  given 
as  in  Fig.  48,  and  let  a"  be  the  vertical  projection  of  a  point  on  its 
surface.  B"  will  be  the  vertical  projection  of  two  elements  pass- 
ing through  a;  B""  and  B,*"  will  be  the  horizontal  projections  of 
these  elements,  and  a*  and  a,*  will  the  horizontal  projections  of 
the  two  points  vertically  projected  at  a". 

Second.  To  draio  a  plane  tangent  to  the  cone  at  the  point  cC  a^. 
—  Here,  as  in  the  case  of  the  cylinder,  H  P,  drawn  tangent  to  the 
base  of  the  cone  at  c\  will  be  the  horizontal  trace  of  the  tangent 
plane  (Art.  79),  and  the  trace  of  the  horizontal  Y  determines  one 
point  s''  in  V  P.  V  P  could  have  been  determined  by  finding  the 
vertical  trace  (/"  of  the  element  of  contact  B. 

V  Q  and  H  Q,  found  in  the  same  way,  are  the  traces  of  a  plane 
tangent  to  the  cone  at  the  point  «"«,*. 

85.  Fiff.  49  shows  the  construction  when  the  axis  of  the  cone 
is  parallel  to  both  V  and  H.  Let  n"  be  the  vertical  projection  of 
a  point  on  its  surface.  Pass  the  profile  plane  X  through  the  base 
of  the  cone,  and  revolve  it  about  V  X  into  V.  G"  is  the  vertical 
projection  of  the  elements  through  n.  In  revolution  d"-'  moves  to 
d^  and  d,%  and  in  counter  rovolution  they  are  found  in  plan  at 


26  DESCRIPTIVE   GEOMETRY. 

d^  and  cZ,^ ;  G^  and  G,^  drawn  through  df"  and  c?,*  and  the  vertex 
o\  are  the  horizontal  projections  of  the  two  elements  vertically 
projected  in  G^  and  n^  and  n^  are  the  horizontal  projections 
required. 

In  finding  the  traces  of  the  tangent  plane  through  the  point 
71"  w'',  we  find  one  point  in  each  trace, —  that  is,  c"  and  tw*,  just  the 
same  as  in  Fig.  47 ;  another  point  in  each  trace  is  obtained,  of 
course,  by  finding  the  two  traces  of  the  element  of  contact  G ; 
joining  the  horizontal  and  vertical  traces,  respectively,  we  have 
V  P  and  H  P.     The  plane  Q  is  found  in  the  same  way. 

If  the  cone  had  been  so  situated  that  the  traces  of  the  element 
of  contact  did  not  fall  within  easy  reach,  we  would  have  made 
use  of  another  profile  plane,  getting  two  more  points,  as  we  did 
t'"  and  yn''. 

86.  Problem  25.  —  G-iven  one  projection  of  a  point  on  the  sur- 
face of  a  double  curved  surface  of  revolution  to  find  its  other  pro- 
jection, and  to  p)  ass  a  plane  tangent  to  the  surface  through  this  point. — 
Let  the  surface  be  a  torus,  represented  as  shown  in  Fig.  50  (Arts. 
76  and  77),  and  let  n^  be  the  horizontal  projection  of  a  point  on 
its  surface.  Pass  the  meridian  plane  Y  through  n ;  H  Y,  drawn 
through  n^  and  A\  is  its  horizontal  trace  ;  revolve  this  plane  into 
the  position  of  the  principal  meridian  plane ;  n^  revolves  in  the 
arc  of  a  circle  to  6'\  which,  being  in  the  principal  meridian,  is  ver- 
tically projected  at  s"'  and  s," ;  in  counter  revolution  s"  and  s^  move 
to  7f  and  11^  vertically  above  w*.  Hence  w"  and  n^  are  the  verti- 
cal projections  of  two  points  on  the  surface  of  the  torus  horizon- 
tally projected  at  w\ 

Second.  To  draw  a  plane  tangent  to  the  torus  at  the  point  n.  — 
This  plane  must  contain  a  line  tangent  to  the  surface,  and  be  per- 
pendicular to  a  normal  at  that  point  (Art.  81).  T/',  drawn  tan- 
gent to  the  principal  meridian  at  s",  will  be  the  revolved  position 
of  the  tangent  line,  and  N/,  drawn  through  s"  and  the  centre  of 
the  generating  circle,  the  revolved  position  of  the  normal.  In 
counter  revolution  the  points  m  and  o  do  not  move,  being  in  the 
axis ;  N/  moves  to  W,  and  T/  to  T"".  Both  the  tangent  and  nor- 
mal, since  they  are  in  the  meridian  plane,  are  horizontally  pro- 
jected in  H  Y.     g''  is  the  horizontal  trace  of  T.    Since  the  tangent 


DESCRIPTIVE  GEOMETRY.  27 

plane  is  to  contain  T  and  be  perpendicular  to  N,  H  P,  drawn 
through  g^  perpendicular  to  N\  and  V  P,  drawn  through  the  point 
where  H  P  intersects  G  L,  perpendicular  to  N'',  are  the  traces  of 
the  tangent  plane. 

A  point  r"  in  V  P  could  have  been  found  by  means  of  the  hori- 
zontal X  of  the  plane  P,  and  still  another  by  finding  the  vertical 
trace  /"  of  the  tangent  line. 


CHAPTER  IV. 

Intersectiok  of  Planes  and  Solids,  and  the  Develop- 
ment OF  Solids. 

87.  To  find  the  intersection  of  any  solid  with  any  secant  plane, 
pass  a  series  of  auxiliary  planes  through  the  solid  and  the  secant 
plane.  They  will  cut  lines  (straight  or  curved)  from  the  solid, 
and  straight  lines  from  the  secant  plane,  the  intersections  of  which 
are  points  of  the  required  curve  of  intersection. 

This  is  a  general  statement  applicable  alike  to  prisms,  pyra- 
mids, cylinders,  cones,  or  double  curved  surfaces  of  revolution. 
While  the  auxiliary  planes  may  be  taken  in  any  position,  yet,  for 
simplicity,  they  should  be  chosen  in  such  a  position  as  to  cut  the 
simplest  curve  from  the  solid, —  that  is,  straight  lines  or  circles  if 
possible. 

In  case  of  any  solid  having  rectilinear  elements,  as  the  cylin- 
der, cone,  prism,  pyramid,  etc.,  the  above  process  is  practically 
the  same  as  finding  where  a  certain  number  of  elements,  or  edges, 
pierce  the  secant  plane  (Arts.  56,  57,  and  58),  since  auxiliary 
planes  may  be  chosen  so  as  to  cut  elements,  or  edges,  from  the 
solid. 

88.  The  true  size  of  the  section  can  always  be  found  by  revolv- 
ing it  about  the  trace  of  the  secant  plane  into,  or  parallel  to,  one 
of  the  coordinate  planes  (Arts.  45,  46,  and  47). 

89.  The  tangent  line  to  the  curve  of  intersection  at  any  point 
is  the  intersection  of  the  plane  tangent  to  the  surface  at  that 
point  with  the  secant  plane ;  for  the  line,  to  be  tangent  to  the 
curve,  must  lie  in  the  plane  of  the  curve, —  that  is,  the  secant 
plane,  and  also  in  the  plane  tangent  to  the  surface  at  that  point ; 
hence  at  their  intersection. 


DESCRIPTIVE   GEOMETRY.  29 

90.  To  develop  a  body  is  to  find,  on  a  flat  surface,  the  true 
size  and  shape  of  the  surface,  or  covering,  of  the  object.  Single 
curved  surfaces  and  solids  bounded  by  planes  can  be  developed ; 
double  curved  surfaces  cannot  be,  except  approximately. 

Solids  bounded  by  planes  are  developed  by  finding  the  true 
size  and  shape  of  each  successive  face. 

Single  curved  surfaces,  as  the  cylinder,  cone,  etc.,  are  developed 
by  placing  one  element  in  contact  with  a  plane,  and  rolling  the 
solid  until  every  element  has  touched  this  plane.  That  portion 
of  the  plane  covered  by  the  solid  in  its  revolution  is  the  develop- 
ment of  the  solid.  When  a  curved  surface  is  developed,  any 
curved  line  upon  it  will  be  developed  into  some  curve,  whose 
length  will  be  equal  to  that  of  the  original  curve. 

CYLINDERS. 

91.  To  find  the  intersection  of  any  cylinder  with  an  oblique 
plane,  assume  the  auxiliary  planes  so  that  they  will  cut  elements 
from  the  cylinder,  and  the  general  statement  for  the  intersection 
of  any  solid  with  a  plane  (Art.  87)  becomes  for  the  cylinder  as 
follows :  Pass  a  series  of  auxiliary  planes  through  the  cylinder 
parallel  to  its  axis  and  perpendicular  to  one  of  the  coordinate 
planes.  Each  auxiliary  plane  cuts  two  elements  from  the  cylin- 
der (except  the  two  tangent  planes  which  cut  but  one  each)  and 
a  right  line  from  the  secant  plane.  The  intersections  of  this  line 
and  elements  give  two  points  of  required  curve. 

92.  Problem  26.  — To  find  the  intersection  of  a  right  cylinder 
ivith  a  circular  base  with  an  oblique  plane,  to  draw  a  tangent  to  this 
curve  at  any  point,  to  find  the  true  size  of  the  section,  to  develop  the 
cylinder,  to  trace  upon  the  development  the  curve  of  iyiter section,  and 
to  draw  a  tangent  to  the  developed  curve.  —  Let  the  cylinder  be 
located  as  shown  in  Fig.  51,  and  P  be  the  secant  plane.  The  auxil- 
iary plane  X  is  parallel  to  the  axis  and  perpendicular  to  H 
(Art.  91) ;  it  cuts  the  two  elements  C  and  D  from  the  cylinder 
and  the  line  X  from  P  ;  X  intersects  C  at  a,  and  D  at  b,  which 
are  two  points  of  the  curve.  Other  points  could  be  determined 
in  the  same  way. 

The  rest  of  the  points  were  found  by  assuming  the  auxiliary 
planes  as  horizontal,  which  may  be  a  trifle  shorter,  for  this  case, 


80  DESCRIPTIVE   GEOMETRY. 

than  the  first  method.  The  horizontal  auxiliary  plane  R  cuts  out 
of  the  cylinder  a  horizontal  circle,  and  out  of  the  plane  the  hori- 
zontal line  R ;  this  line  intersects  the  circle  at  the  points  c  and  d^ 
which  are,  therefore,  two  points  of  the  curve.  In  the  same  way 
the  plane  S  gives  the  points  g  and  k,  and  the  plane  Z  the  points 
on  and  ??,  etc. 

To  di'atv  a  tangent  to  the  point  n  in  the  curve. — The  tangent  line 
to  the  curve  at  the  point  n  is  the  intersection  of  the  plane  tangent 
to  the  cylinder  at  that  point  and  the  plane  P  (Art.  89).  V  T 
and  H  T  are  the  traces  of  the  plane  tangent  to  the  cylinder  at  the 
point  n  (Art.  82).  T''  and  T^  the  intersection  of  the  planes  P 
and  T,  are  the  projections  of  the  tangent  line.  Since  the  point  n 
must  be  one  point  in  the  tangent  line,  it  is  only  necessary  to  find 
one  other  point,  as  r,  and  the  tangent  line  is  determined, —  that  is, 
it  is  not  necessary  to  find  the  vertical  trace  of  the  tangent  plane, 
which  shortens  the  construction. 

To  find  the  true  size  of  the  section. — Each  point  in  the  curve, 
as  a,  c,  g.,  m,  b,  etc.,  revolve  to  a,,  c,,  g,,  m,,  J,,  etc.  (Art.  45),  and 
the  true  size  of  the  section  is  shown. 

The  tangent  line  T  revolves  with  the  section.  The  point  r* 
does  not  move,  since  it  lies  in  the  axis  H  P,  and  the  point  n 
revolves  to  n, ;  hence  T  revolves  to  T, . 

To  develop  the  cylinder. — If  we  place  one  element  of  the  cylin- 
der, as  D,  Fig.  52,  on  a  plane,  and  roll  the  cjdinder  until  the  same 
element  coincides  with  the  plane  again,  that  part  of  the  plane 
covered  in  this  revolution  will  be  the  development  (Art.  90). 
The  element  will,  of  course,  be  seen  in  its  true  length,  and  a  right 
section,  being  perpendicular  to  the  elements,  unrolls  in  a  straight 
line  perpendicular  to  the  elements,  and  equal  in  length  to  the 
circumference  of  the  right  section.  Hence  h!"  /i,\  nj"  k\  kj'  d\  etc., 
Fig.  52,  equal  to  5*  >A  w*/:;\  k^'d",  etc..  Fig.  51,  laid  off  on  a 
straight  line  will  be  the  development  of  the  base,  and  perpendicu- 
lars erected  at  these  points  will  be  the  position  of  the  several  ele- 
ments; hi"  s  is  made  equal  to  the  height  of  the  cylinder,  and  st^ 
drawn  parallel  to  hj'  bf\  completes  the  development  of  the  whole 
cylinder. 

To  trace  upon  the  development  the  curve  of  intersection.  —  Make 
6,*  J,",  w,*  w,",  k,'' k",  etc.,  equal  to  the  heights  of  the  points  6,  n,  k, 
etc.,  of  the  curve  above  H.     The  curve  traced  througli  these  sue- 


DESCRIPTIVE  GEOMETRY.  31 

cessive  points  will  be  the  development  of  the  curve  of  intersec- 
tion. 

To  draw  the  tmigent  line  to  the  j^oint  n  on  the  development.  — 
The  point  w,"  will  be  one  point  of  the  tangent  line  sought.  The 
point  r  being  in  the  plane  of  the  base  must  be  found  in  the  line 
on  which  the  base  unrolls,  and  at  a  distance  from  nj"  equal  to  the 
distance  of  the  point  r  from  the  foot  of  the  element  along  which 
the  plane  is  tangent, —  that  is,  make  nj"  r^  equal  to  n^  r''  and  join- 
ing r/'  and  n'^  we  have  the  developed  position  of  the  tangent  line. 
This  line  must  be  tangent  to  the  developed  position  of  the  curve. 

93.  Problem  27. — To  find  the  intersection  of  an  oblique  cylin- 
der with  any  oblique  plane,  to  draw  a  tangeiit  to  any  point  of  the 
curve,  and  to  develop)  that  part  of  the  cylinder  betweerc  the  base  and 
the  secant  plane,  —  Let  the  cylinder  be  given  as  in  Fig.  53,  and 
let  P  be  the  cutting  plane. 

Assume  the  auxiliary  planes,  R,  S,  U,  etc.,  parallel  to  the  axis 
and  perpendicular  to  H  (Art.  91).  The  plane  U  cuts  two  ele- 
ments, F  and  N,  from  the  cylinder  and  the  line  U  from  the  plane  ; 
X  and  w,  the  intersections  of  this  line  with  these  elements,  are  two 
points  of  the  curve.  Similarly  the  plane  R  gives  the  point  o ;  S 
the  points  n  and  p ;  W  gives  I  and  r ;  X  gives  k  and  s ;  Y  gives 
g  and  t;  and  Z  the  point  /,  In  the  construction  of  the  different 
lines  of  intersection  R,  S,  U,  etc.,  it  is  only  necessary  to  construct 
one  of  the  lines,  the  others  are  drawn  through  the  vertical  projec- 
tions of  the  horizontal  traces,  parallel  to  the  first,  since  the  lines 
of  intersection  of  several  parallel  planes  with  an  oblique  plane 
must  be  parallel  to  each  other. 

The  tangent  line  is  found  as  explained  in  the  last  problem.  T 
is  the  plane  tangent  to  the  cylinder  along  the  element  through  t, 
and  T""  and  T'\  the  line  of  intersection  of  the  planes  T  and  P,  are 
the  projections  of  the  tangent  line  to  the  curve  at  the  point  t. 

To  develop  that  p)<^rt  of  the  cylinder  between  the  base  and  the 
secant  p)lane.  —  Since  neither  the  plane  of  the  base  nor  the  plane 
of  the  section  just  found  is  perpendicular  to  the  axis,  the  sections 
are  not  right  sections,  hence  they  will  not  unroll  in  straight 
lines.  Consequently  the  first  step  necessary  in  the  solution  of 
this  part  of  the  problem  is  to  find  a  right  section.  K""  K^,  the 
section  cut  by  the  plane  Q,  which  is  perpendicular  to  the  axis,  is 


32  DESCRIPTIVE   GEOMETRY. 

a  right  section ;  K'  is  the  true  size  of  this  section ;  and  K, ,  Fig. 
54,  is  its  development,  its  length  being  equal  to  the  periphery  of 
the  section  K'.  Find  the  true  lengths  of  the  successive  elements 
between  this  right  section  and  the  base,  and  lay  them  off  perpen- 
dicular to  the  line  K,  on  lines  which  represent  the  developed  posi- 
tion of  the  several  elements.  The  curved  line  A,  A,  drawn 
through  the  points  thus  found  is  the  development  of  the  peri- 
phery of  the  base. 

To  develop  the  section  cut  out  by  the  plane  P,  find  the  true 
lengths  of  the  successive  elements  between  this  section  and  the 
base,  laying  them  off  from  the  curved  line  A,  A,,  and  the  curved 
line  I,  m,  n,  etc.,  is  the  development  of  this  section.  This  curve 
could  also  have  been  found  by  finding  the  true  lengths  of  the  ele- 
ments between  the  right  section  and  the  section  cut  by  P,  and 
laying  these  lengths  off  from  the  development  K,  of  the  right 
section.  Hence  the  surface  included  between  the  curved  lines 
A,  A,  and  l,m,n,  etc.,  is  the  development  of  that  portion  of  the 
cylinder  between  the  plane  P  and  the  base. 

To  draiv  the  tangent  line  to  the  point  t  on  the  development. — The 
point  e  is  at  a  certain  known  distance  from  the  point  f,  and  also  a 
certain  other  known  distance  from  the  point  iv ;  hence  with  w, , 
Fig.  54,  as  a  centre  and  a  radius  equal  to  w^  e^  Fig.  53,  describe 
an  arc ;  with  t^.  Fig.  54,  as  a  centre  and  a  radius  equal  to  the  true 
length  of  the  line  t  g.  Fig.  53,  describe  another  arc ;  the  point  e, , 
where  these  two  arcs  intersect,  will  be  one  point  of  the  developed 
position  of  the  line.  t,  must  be  another,  therefore  T, ,  drawn 
through  t,  and  e,,  is  the  tangent  sought. 

CONES. 

94.  The  general  statement  for  the  intersection  of  any  cone 
witli  any  plane  is  as  follows :  Pass  a  series  of  auxiliary  planes 
through  the  vertex  of  the  cone,  and  perpendicular  to  one  of  the 
coordinate  planes.  These  planes  cut  elements  from  the  cone  and 
lines  from  the  secant  plane,  the  intersections  of  which  give  points 
on  the  curve. 

95.  Problem  28. — To  find  the  intersection  of  a  right  cone  ivith 
a  circular  hhse  ivith  an  oblique  plane,  etc.,  as  in  Problem  26.  —  Let 
the  cone  and  cutting  plane  P  be  given  as  in  Fig.  55.     The  auxil- 


DESCmrTIVE   GEOMETRY.  33 

iary  plane  R  passes  through  the  vertex  and  is  perpendicular  to  H 
(Art.  94) ;  it  cuts  the  two  elements  A  and  B  from  the  cone  and 
the  line  R  from  P  ;  R  intersects  A  at  c,  and  B  at  g^  which  are  two 
points  of  the  curve.  These  two  points  are  the  highest  and  lowest 
points  of  the  curve,  since  the  line  R  is  the  line  of  greatest  decliv- 
ity of  the  plane  P.  Other  points  could  be  determined  in  the 
same  way. 

The  rest  of  the  points  were  found  by  assuming  horizontal  auxil- 
iary planes  X,  Y,  and  Z.  The  plane  X  cuts  out  of  the  cone  a 
horizontal  circle,  and  out  of  the  plane  the  horizontal  line  X ;  this 
line  intersects  the  circle  at  the  points  d  and  p,  which  are,  there- 
fore, two  points  of  the  curve.  In  the  same  way  the  plane  Y  gives 
the  points  e  and  w,  and  the  plane  Z  the  points  /  and  m. 

The  tangent  line  T,  and  the  true  size  of  the  section,  are  found 
in  the  same  way  as  described  for  the  cylinder  in  Problem  26. 

To  develop  the  cone  and  trace  on  the  development  the  curve  of 
intersection.  —  In  a  right  cone  all  the  elements  pass  through  the 
vertex,  and  are  of  the  same  length,  hence  the  vertex  remains 
stationary  and  the  base  unrolls  in  the  arc  of  a  circle,  of  which  the 
vertex  is  the  centre,  and  the  slant  height  of  the  cone  (that  is,  the 
true  length  of  the  element)  is  the  radius.  The  length  of  this  arc 
must  be  equal  to  the  circumference  of  the  base.  TherefoKe,  with 
o\  Fig.  bQ.,  as  a  centre,  and  a  radius  equal  to  the  true  length  of  an 
element,  describe  an  arc  D'  equal  in  length  to  the  circumference 
of  the  base,  and  o'  t'  T>'  t'  is  the  development  of  the  surface  of  the 
cone.  Particular  elements  may  be  drawn  on  the  development  by 
laying  off  distances  on  the  arc  D'  equal  to  the  distance  apart  of 
the  elements  at  the  base  of  the  cone,  and  joining  these  points  with 
the  vertex  o'.  To  find  the  development  of  the  curve  of  intersec- 
tion, lay  off  from  o\  Fig.  56,  on  its  corresponding  element  the  true 
length  of  the  element  from  the  vertex  to  the  section,  and  joining 
the  points  thus  found,  g\  /',  e',  etc.,  gives  the  development 
sought. 

To  draw  the  tangent  to  the  development  at  the  'point  ^:>.  —  The 
line  C,  tangent  to  the  base  of  the  cone,  is  perpendicular  to  the 
element  s  o,  hence  it  will  be  so  after  development.  Therefore 
draw  C,  Fig.  56,  perpendicular  to  s'  o',  and  lay  off  on  it  s'  r'  equal 
to  **  r\  Fig.  55;  r  is  also  a  point  of  the  tangent  line  T,  hence  r' 
must  be    one  point  of   its  developed  position,   and  p'  must  be 


34  DESCRIPTIVE   GEOMETRY. 

another  point,  consequently  T'  drawn  through  r'  and  f'  is  the 
tangent  required. 

96.  Fig.  57  shows  the  construction  when  the  axis  of  the  cone 
is  parallel  to  both  V  and  H.  The  method  of  finding  the  point  o," 
is  explained  in  the  next  article.  Fig.  58  shows  the  development 
of  the  cone  and  the  tangent  line. 

97.  Problem  29.  — To  find  the  intersection  of  an  oblique  cone 
with  any  oblique  plane^  to  draw  a  tangent  to  any  point  of  the  curve^ 
and  to  develop  that  ptortion  of  the  cone  between  the  base  and  the 
secant  plane.  —  Let  the  cone  be  given  as  in  Fig.  59,  and  let  P  be 
the  cutting  plane. 

Assume  the  auxiliary  planes  R,  S,  X,  etc.,  through  the  vertex 
and  perpendicular  to  H  (Art.  94).  The  plane  S  cuts  two  ele- 
ments, A  and  B,  from  the  cone,  and  the  line  S  from  the  plane ; 
g  and  r,  the  intersections  of  this  line  with  these  elements,  are 
two  points  of  the  curve.  Similarly  the  plane  R  gives  the  point 
/;  X  the  points  k  and  p ;  Y  the  points  I  and  n ;  and  Z  the 
point  m. 

In  the  construction  of  the  different  lines  of  intersection  R.,  S, 
X,  etc.,  it  is  not  necessary  to  go  through  the  whole  process  of 
finding  each  of  them.  In  the  case  of  the  cylinder,  these  lines  of 
intersection  were  parallel,  here  they  intersect  in  a  common  point 
0,.  The  auxiliary  planes,  being  all  perpendicular  to  H,  and  pass- 
ing through  the  vertex  of  the  cone,  will  intersect  each  other  in  a 
common  line,  which  must  be  perpendicular  to  H,  and  contain  the 
vertex;  D''  will  be  the  vertical  projection  of  this  line,  and  o*  its 
horizontal  projection ;  hence  the  point  o,  must  be  on  this  line, 
which  is  common  to  all  the  auxiliary  planes,  and  on  the  plane  P, 
therefore  at  the  point  where  they  intersect.  o^  is  the  hoiizontal 
projection  of  this  point,  and  o/,  constructed  as  explained  in  Prob- 
lem 5,  is  its  vertical  projection.  This  point  could  be  found  by 
extending  one  line  of  intersection,  found  regularly,  until  it  inter- 
sects the  vertical  line  joining  the  projections  of  the  vertex. 

The  tangent  line  T  at  the  point  ?,  the  true  size  of  the  section, 
etc.,  are  found  as  explained  in  a  previous  problem. 

To  develop  that  part  of  the  cone  which  is  between  the  base  and 
the  secant  plane.  —  Since  the  vertex  remains  stationary  we  have  a 
means  of  locating  tlie  developed  positions  of  the  elements  with 


DESCRIPTIVE   GEOMETRY.  35 

respect  to  each  other  without  getting  a  right  section  as  in  the 
cylinder. 

Draw  any  right  line  o' t\  Fig.  60,  and  on  it  lay  off  the  true 
length  of  any  one  of  the  elements,  as  E,  Fig.  59.  Find  the  true 
length  of  the  next  element  A,  with  this  length  as  a  radius  and  o' 
as  a  centre,  describe  an  indefinite  arc.  The  distance  these  two 
elements  are  apart  at  their  lower  extremities  must  be  equal  to  the 
length  of  that  part  of  the  circumference  of  the  base  included 
between  ^-  and  d/".  Hence  with  t'  as  a  centre,  and  a  radius  equal 
to  i''c?^  describe  an  arc  cutting  the  other  arc  at  d' ;  join  d'  with 
o',  and  we  have  the  developed  position  of  the  element  A.  The 
other  elements  are  developed  in  the  same  way.  The  greater  the 
number  of  elements  developed,  the  greater  will  be  the  accuracy 
of  the  development. 

To  develop  the  section  cut  out  by  the  plane  P,  find  the  true 
lengths  of  each  element  from  the  vertex  to  the  section  o/,  o  g, 
etc.,  and  lay  them  off  from  o'  on  the  developed  positions  of  those 
elements  respectivel}^  o'f\  o' g\  etc.  That  portion  of  the  surface 
included  between  the  development  of  the  base  t'  c'  x'  d' t\  and  that 
of  the  section /''^' ^'  etc.,  will  be  the  development  of  that  part  of 
the  cone  between  the  base  and  the  secant  plane  P. 

The  tangent  line  to  the  point  I  is  developed  as  explained  in 
Problem  27,  the  point  e  being  in  ^*  and  T"",  and  at  a  known  dis- 
tance from  the  points  x  and  I. 

DOUBLE  CURVED  SURFACES  OF  REVOLUTION. 

98.  The  general  statement  for  the  intersection  of  a  double 
curved  surface  of  revolution  with  any  plane  is  as  follows :  Pass  a 
series  of  auxiliary  planes  through  the  solid  perpendicular  to  its 
axis.  These  planes  cut  circles  from  the  solid  and  lines  from  the 
secant  plane,  the  intersections  of  which  give  points  on  the 
curve. 

99.  Problem  30. — To  find  the  intersection  of  a  double  curved 
surface  of  revolution  with  an  oblique  plane,  and  to  draw  a  tangent 
to  the  curve  at  any  point.  — Let  the  double  curved  surface  of  revo- 
lution be  a  torus  given  as  in  Fig.  61,  and  let  P  be  the  cutting 
plane.  Assume  the  auxiliary  planes  R,  S,  X,  etc.,  perpendicular 
to  the  axis.     The  plane  Y  cuts  from  the  torus  two  circles,  hori- 


86  DESCRIPTIVE  GEOMETRY. 

zontallj  projected  in  A''  and  B\  and  vertically  projected  in  V  Y, 
it  cuts  from  the  plane  P  the  horizontal  line  Y ;  cZ,  /,  ?,  and  w,  the 
intersections  of  these  circles  and  line,  are  four  points  on  the 
curve.  Similarly  the  plane  R  gives  the  point  a;  the  plane  X 
gives  c,  g,  k,  and  o  \  the  plane  S  gives  h  and  'p ;  and  the  plane  Z 
gives  e  and  m. 

The  tangent  line  T  to  any  point  as  w,  and  the  true  size  of  the 
section,  etc.,  are  shown  in  the  iigure,  and  need  no  further  ex- 
planation. 

SOLIDS   BOUNDED   BY  PLAIN   SURFACES. 

100.  In  finding  the  intersection  of  prisms,  pyramids,  or  any 
solid  bounded  by  plane  surfaces  with  an  oblique  plane,  pass  auxil- 
iary planes  through  the  edges  of  the  solid,  and  perpendicular  to 
V  or  H  ;  or,  in  other  words,  find  where  each  edge  of  the  solid 
pierces  the  secant  plane  by  Art.  56. 

101.  Problem  31.  — To  find  the  intersection  of  a  prism  ivith  an 
oblique  plane.  —  Let  the  prism  be  pentagonal,  and  given  as  in 
Fig.  62,  and  let  P  be  the  cutting  plane.  The  edges  A,  B,  G, 
etc.,  pierce  the  plane  P  in  the  points  a,  h,  c,  etc. ;  joining  these 
points  we  get  the  two  projections  of  the  intersection.  Its  true 
size  is  shown  at  a,  5,  c,  d,  e, . 

The  development  may  be  found  directly  here,  without  getting  a 
right  section  as  in  the  cylinder,  by  revolving  each  face  about  its 
edge  resting  on  H  into  H  (Art.  47).  a  6^/ revolves  to  f"  g''  on-, 
h  ck g  to  g^-  k^  r p.,  etc. 

102.  We  have  seen  how,  if  we  have  a  solid  given  by  its  pro- 
jections, we  can  obtain  its  covering  or  development;  hence,  if  we 
have  the  covering  given,  we  are  able  to  construct  the  projections 
of  the  solid. 

103.  Problem  82. — To  construct  the  projections  of  a  solid  from 
its  covering^  and  find  the  intersection  of  this  solid  with  an  oblique 
plane.  —  Let  the  covering  consist  of  eight  regular  hexagons  and 
six  squares,  disposed  as  shown  in  Fig.  63.  Let  one  of  the  hexa- 
gons be  considered  as  the  base,  and  let  it  be  placed  parallel  to  H,  so 
that  one  of  its  sides,  a  6,  is  perpendicular  to  V.  Revolve  the  hexa- 
gon A  and  the  square  K  until  the  sides  a''g'  and  a''  k'  coincide,  the 
extremities  g'  and  k'  of  tiiese  sides  will  meet  at  a  point  in  space, 


DESCRIPTIVE  GEOMETRY.  87 

of  winch  the  horizontal  projection  is  evidently  a^ ;  revolve 
square  M  until  its  side /'T  coincides  with  the  side  o''/'' of  the 
revolved  hexagon  A,  the  extremities  meet  at  a  point  in  space  of 
which  the  horizontal  projection  is  o  h.  It  is  evident  that  the 
horizontal  projections  of  the  two  squares  K  and  M,  in  their 
revolved  positions,  are  a''  V'  s^  x^  and  e;*/^  o'^p''.  To  complete  the 
horizontal  projection  of  the  hexagon  A,  we  draw  through  jt^  the 
line  a^  w''  parallel  and  equal  to  the  line/*o'';  through  w''  the  line 
w*w*  equal  and  parallel  to  aJ'f^;  and  finally  join  n''  and  o^,  and 
w^o*  will  be  equal  and  parallel  to  a^'  ji^  (Art.  13).  The  vertical 
projection  of  the  base  is  the  line  a"/"  e"  at  any  required  height 
above  H.  The  vertical  projection  of  the  square  K,  in  its  revolved 
position,  is  the  line  a" a;",  found  as  follows:  Since  the  plane  of  the 
square  is  perpendicular  to  V,  the  point  k'  moves  in  revolution  in 
the  arc  of  a  vertical  circle,  with  the  point  a  as  a  centre ;  hence, 
when  the  point  has  gone  to  a/*,  its  vertical  projection  must  be  at 
a;",  where  a  perpendicular  through  a^  intersects  this  circle.  Since 
the  squares  K  and  M  are  inclined  at  equal  angles  to  H,  their  top 
points  must  be  the  same  height  above  H  in  their  revolved  posi- 
tions ;  hence  o"  and  p"  must  be  found  in  a  horizontal  line  through 
x'\  and  vertically  above  o''  and  jt?^.  Hence  the  vertical  projection 
of  the  square  M  is  e^'f  d" p".  The  vertical  projection  of  the 
hexagon  A  is  a"/"  o"  w"  w"  a;^  the  points  a^/^  o",  and  x"  being  co- 
incident with  those  of  the  squares  already  found.  The  point  w" 
is  obtained  by  dropping  a  perpendicular  from  w\  and  producing 
it  until  it  meets  a  line  drawn  through  a;"  parallel  to  f  d" ;  or, 
in  other  words,  make  a;"?^"  equal  and  parallel  to/^o";  make  g"  n" 
equal  and  parallel  to  a'" a;";  and  n" w"  eqvial  and  parallel  to  a''/". 

The  other  hexagons  and  squares  are  revolved  after  the  same 
principle,  and  the  completed  projections  are  obtained  as  shown  in 
the  figure. 

To  find  its  intersection  with  the  plane  P.  —  Find  where  the 
edges  of  the  solid  pierce  the  plane  by  Art.  56 ;  some  of  them  will 
and  some  will  not;  hence  assume  any  one  that  you  think  will 
pierce  it,  and  see  if  it  does  by  actual  construction.  If  it  does  not, 
try  another,  and  so  on.  The  edge  ep  pierces  the  plane  P  at  the 
point  6  ;  /o  at  6  ;  wn  at  4,  etc. ;  giving  12345678  as  the  sec- 
tion.    Its  true  size  is  found  as  usual. 

The  lines  1,  2,,  2,  3,,  3,  4,,  4,  5,,  etc.,  are  the  developed  posi- 
tions of  the  lines  cut  from  the  several  faces  by  the  secant  plane. 


CHAPTER  V. 

llSTTERSECTIOIf   OF   SOLIDS. 

104.  To  find  the  curve  of  intersection  of  any  two  solids  what- 
ever, pass  a  series  of  auxiliary  surfaces  (usually  planes)  through 
the  two  solids.  They  will  cut  lines  (straight  or  curved)  from 
each  solid,  the  intersections  of  Which  are  points  of  the  required 
curve. 

While  the  auxiliary  planes  may  be  taken  in  any  position,  j'et, 
for  convenience,  they  should  be  chosen  so  as  to  cut  the  simplest 
curves  from  the  solid, —  i.e.,  straight  lines  or  circles. 

105.  The  following  is  a  list  of  special  cases  showing  what 
auxiliary  planes  should  be  taken  to  cut  the  simplest  curves  :  — 

First.  Two  cylinders  with  a?tes  oblique  to  each  other ;  choose 
auxiliary  planes  parallel  to  both  axes.  Each  plane  cuts  elements 
from  each  solid. 

Second.  Two  cones  with  axes  perpendicular  or  oblique  to  each 
other;  choose  auxiliary  planes  passing  through  a  line  containing 
the  vertex  of  each  cone.  Each  plane  cuts  elements  from  each 
solid. 

Third.  Cylinder  and  cone  with  axes  oblique  to  each  other; 
choose  auxiliary  planes  passing  through  vertex  of  cone  and  par- 
allel to  axis  of  cylinder.  Each  plane  cuts  elements  from  each 
solid. 

Fourth.  In  case  of  the  intersection  of  cylinders,  cones,  double 
curved  surfaces  of  revolution,  prisms,  and  pyramids  (any  one 
with  one  of  the  same  or  different  kind)  with  their  axes  parallel ; 
clioose  auxiliary  planes  perpendicular  to  the  axes.  The  planes 
will  cut  from  these  solids  simple  figures  (from  the  cylinder,  cone, 
and  double  curved  surface  of  revolution,  circles,  and  from  the 


DESCRIPTIVE   GEOMETRY.  39 

prisms  and  pyramids,  polygons  similar  to  their  bases).  One  of 
the  coordinate  planes  should  be  taken  perpendicular  to  the  axes. 

Fifth.  In  case  of  two  cylinders,  cylinder  and  double  curved 
surface  of  revolution,  cylinder  and  cone,  whose  axes  are  perpen- 
dicular to  each  other ;  choose  auxiliary  planes  perpendicular  to 
one  cylinder  in  first  case,  to  axis  of  double  curved  surface  of  revo- 
lution in  second  case,  to  axis  of  cone  in  third  case.  Coordinate 
planes  should  be  taken  perpendicular  to  the  axes  respectively. 

Sixth.  In  case  of  two  double  curved  surfaces  of  revolution 
with  axes  oblique  and  intersecting,  pass  a  series  of  auxiliar}^ 
spheres,  whose  centres  are  at  the  intersection  of  the  axes,  through 
the  two  solids. 

A  prism  can  be  substituted  in  place  of  the  cylinder,  and  pyra- 
mid in  place  of  the  cone,  in  any  of  the  above  cases. 

106.  The  tangent  line  to  the  curve  of  intersection  at  any  point 
is  the  line  of  intersection  of  the  planes  tangent  to  each  solid  at 
that  point. 

197.  Problem  33. — To  find  the  curve  of  intersection  of  tivo 
cylinders  whose  axes  are  oblique  to  each  other,  and  to  draw  a  line 
tangent  to  any  point  in  the  curve. — Let  the  cylinders  be  given  as  in 
Fig.  64.  This  comes  under  the  first  special  case  (Art.  105),  and 
the  auxiliary  planes  are  taken  parallel  to  the  axis  of  each  cylin- 
der. To  find  the  horizontal  trace  of  such  a  plane  (the  vertical 
trace  is  not  needed)  we  have  simply  to  pass  a  plane  through  one 
axis  and  parallel  to  the  other  (Art.  63).  H  S  is  the  horizontal 
trace  of  such  a  plane.  H  R,  H  X,  etc.,  drawn  parallel  to  H  S  will 
be  the  horizontal  traces  of  planes  which  are  parallel  to  both  axes. 
Each  of  these  planes  cuts  elements  from  each  cylinder,  the  points 
of  intersection  of  which  are  points  of  the  required  curve. 

The  plane  S  cuts  the  two  elements  A  and  B  from  the  left  cylin- 
der, and  C  and  D  from  the  right  cylinder  ;  these  elements  inter- 
sect at  the  points  2,  8,  10,  and  16,  which  are  four  of  the  required 
points.  The  plane  R  being  tangent  to  one  of  the  cylinders  only 
gives  two  points,  1  and  9.  The  plane  X  gives  the  four  points, 
3,  7,  11,  and  15.  By  repeating  the  process  any  number  of  points 
can  be  found. 

After  finding  a  number  of  points  on  the  curve,  crossing  and 
recrossing  as  it  is  liable  to  do,  unless  some  system  of  numbering 


40  DESCRIPTIVE  GEOMETRY. 

is  followed,  it  is  very  difficult  to  connect  the  points  so  as  to  form 
the  correct  curve.  An  examination  of  the  system  used  in  figures 
64,  65,  and  68  will  suffice  without  explanation ;  the  same  method 
is  used  whether  the  solids  are  two  cylinders,  two  cones,  or  cylin- 
der and  cone. 

To  determine  whether  there  tvill  he  one  or  two  curves. —  Draw  the 
two  extreme  planes  II  and  Z,  each  of  them  will  be  tangent  to  one 
(but  not  the  same)  cylinder,  and  secant  to  the  other,  hence  there 
will  be  but  one  curve. 

If  the  two  extreme  planes  were  tangent  to  the  same  body,  and 
secant  to  the  other,  it  is  evident  that  the  first  body  would  pass 
through  the  second,  and  there  would  be  two  separate  curves. 

To  determine  what  points  on  the  curve  are  visible.  —  A  point  on 
the  curve  is  visible  in  plan  or  elevation,  if  it  is  the  intersection 
of  two  elements,  both  of  which  are  visible  in  plan  or  elevation. 
Thus  the  point  9,  being  at  the  intersection  of  two  elements  which 
are  visible  in  plan,  must  be  on  a  visible  portion  of  the  curve  in 
plan ;  these  elements  being  invisible  in  elevation,  the  curve  at 
that  point  will  be  invisible  in  elevation.  The  two  elements  whose 
intersection  gives  the  point  5  are  both  visible  in  elevation,  but 
only  one  is  visible  in  plan,  therefore  the  point  is  on  a  portion  of 
the  curve  which  is  visible  in  elevation,  but  not  in  plan. 

To  draio  a  tangent  line  to  any  point  of  the  curve.  —  The  tangent 
line  to  any  point  of  the  curve  is  the  line  of  intersection  of  two 
planes  tangent  to  the  solids  at  that  point.  H  T  is  the  horizontal 
trace  of  the  plane  tangent  to  the  left  cylinder  at  the  point  14,  and 
H  Q  that  of  the  plane  tangent  to  the  right  cylinder  at  the  same 
point ;  T,  the  line  of  intersection  of  these  two  planes,  is  the  tan- 
gent line  required. 

108.  Problem  34. — To  find  the  curve  of  intersection.,  etc.,  of 
tivo  cones  with  axes  oblique  to  each  other.  —  Let  the  cones  be  given 
as  in  Fig.  65.  This  comes  under  the  second  special  case  (Art. 
105),  and  the  auxiliary  planes  are  passed  through  a  line  contain- 
ing the  vertex  of  each  cone. 

The  line  A  passes  through  each  vertex,  and  a^  is  its  horizontal 
trace  ;  hence  any  line,  as  H  X,  H  Y,  etc.,  drawn  through  a\  will 
be  the  horizontal  trace  of  a  plane  which  will  pass  through  the 
vertex  of  each  cone,  and,  consequently,  will  cut  elements  from 


DESCRIPTIVE   GEOMETRY.  41 

them.  H  Y  cuts  the  elements  B  and  C  from  the  left  cone,  and 
D  and  E  from  the  right  cone ;  these  elements  intersect  at  the 
points  2,  4,  6,  and  8,  which  will  be  points  on  the  required  curve. 
Any  number  of  points  may  be  found  in  the  same  way. 

Here,  as  in  the  case  of  the  cylinder  in  the  last  problem,  the  two 
extreme  auxiliary  planes  X  and  Z  are  each  tangent  to  one  (but 
not  the  same)  body  and  secant  to  the  other ;  hence  there  will  be 
but  one  curve. 

In  this  case  one  of  the  cones  is  supposed  to  have  been  removed, 
consequently  those  portions  of  the  curve  will  be  visible  if  they 
are  on  visible  portions  of  the  remaining  cone. 

T  is  the  tangent  to  the  point  w,  and  is  found  exactly  as  in  the 
last  problem. 

109.  Problem  35. — To  find  the  curve  of  intersection^  etc.,  of 
two  cones  when  their  axes  are  perpendicular  to  each  other.  —  Let 
the  cones  be  given  as  in  Fig.  66,  the  axis  of  one  being  perpendicular 
to  H,  and  of  the  other  parallel  to  both  V  and  H.  In  this  case 
we  pass  the  auxiliary  planes  through  a  line  containing  each  ver- 
tex, as  in  the  last  problem,  but  here  it  is  necessary  to  use  a  pro- 
file plane  and  make  use  of  the  traces  of  the  auxiliary  planes  on. 
this  plane,  as  well  as  their  traces  on  H. 

The  horizontal  trace  of  the  line  A,  containing  the  two  vertices 
of  the  cones,  is  a*;  its  trace  on  the  profile  plane  is  b.  H  S,  H  X, 
etc.,  are  the  horizontal  traces  of  planes  which  will  cut  ele- 
ments from  the  two  cones;  6^Z*,  ^''c^,  etc.,  are  their  traces  on 
the  profile  plane.  Revolving  the  profile  plane  about  its  vertical 
trace  into  V,  the  point  b  revolves  to  J,",  and  the  traces  of  the 
auxiliary  planes  on  the  profile  plane  are  shown  at  V  S',  V  X',  etc. 
The  plane  S  cuts  the  element  D  from  the  cone  whose  axis  is  hori- 
zontal, and  B  and  C  from  the  upright  cone ;  the  intersections  of 
these  elements  d  and  e  are  two  points  of  the  curve.  The  plane  Z 
gives  the  points  m,  w,  r,  and  t.  Any  number  of  points  can  be 
found  in  the  same  way. 

Since  the  axes  of  the  two  cones  intersect,  the  curve  will  be  the 
same  on  each  side  of  the  central  plane  Z ;  hence,  since  the  plane 
S  is  tangent  to  one  cone  and  secant  to  the  other,  there  will  be  two 
separate  curves. 

T  is  the  tangent  line  to  a  point  on  the  curve,  being  the  inter- 


42  DESCRIPTIVE   GEOMETRY. 

section  of  the  plane  Q,  which  is  tangent  to  the  cone  whose  axis  is 
horizontal,  and  the  plane  T,  tangent  to  the  other  cone  at  the  com- 
mon point. 

Fig.  67  shows  the  development  of  the  upright  cone,  together 
with  the  holes  in  which  the  horizontal  cone  penetrated  it,  and  the 
tangent  line. 

110.  Problem  36. —  To  find  the  curve  of  intersection,  etc.,  of  a 
cylinder  and  co7ie  with  axes  oblique.  —  Let  the  bodies  be  given  as 
in  Fig.  68.  This  comes  under  the  third  case,  and  the  auxiliary- 
planes  are  passed  through  the  vertex  of  the  cone  parallel  to  the 
axis  of  the  cylinder.  To  do  this,  draw  the  line  A  through  the 
vertex  of  the  cone,  and  parallel  to  the  axis  of  the  cylinder ;  a''  is 
its  horizontal  trace ;  any  line,  as  H  S,  H  X,  etc.,  drawn  through 
this  point  will  be  the  horizontal  trace  of  such  a  plane. 

The  plane  S  cuts  the  element  B  from  the  cone,  and  the  two  ele- 
ments, C  and  D,  from  the  cylinder ;  the  intersections  of  these 
elements  at  1  and  7  are  tw^o  points  of  the  required  curve.  Any 
number  of  other  points  can  be  found  in  the  same  way. 

111.  Problem  37.  —  To  find  the  curve  of  intersection  of  a 
sphere  and  hexagonal  prism.  —  Let  them  be  given  as  in  Fig.  69. 
This  is  an  illustration  of  the  fourth  special  case,  of  which  there 
can  be  a  large  number. 

The  horizontal  plane  X  cuts  the  circle  X  from  the  sphere,  and 
a  hexagon  from  the  prism  ;  these  intersect  at  the  point  n.  Other 
points  are  found  in  the  same  manner. 

112.  Problem  38. — To  find  the  curve  of  intersection  of  a  cylin- 
der and  torus  with  axes  parallel.  —  Let  them  be  given  as  in  Fig. 
70.  This  is  an  illustration  of  the  fifth  special  case.  Here  planes 
should  be  taken  perpendicular  to  the  axis  of  the  torus. 

The  plane  Y  cuts  from  the  torus  the  two  circles  A  and  B,  and  from 
the  cylinder  the  two  elements  C  and  D.  These  circles  and  elements 
intersect  in  the  points  a,  a,,  a„,  a,„,  J,  5,,  all  of  which  are  points  ou 
the  required  curve.    Other  points  are  found  in  tlie  same  manner. 

113.  In  cases  where  the  bodies  are  bounded  by  planes,  and  do 
not  rest  with  their  bases  on  either  of  the  coordinate  planes,  it  is 
easier  to  make  use  of  the  method  described  in  Art.  58  than  to  find 
the  traces  of  the  planes  containing  the  separate  faces  of  the  solid. 


DESCRIPTIVE  geo:metry.  48 

114.  Problem  39.  —  To  find  the  intersection  of  a  triangular 
prism  with  a  triangular  pi/ramid.-"— The  points,  e,f,  and^,  Fig.  71, 
where  the  elements  A,  B,  and  C  of  the  prism  penetrate  the  face  abc 
of  the  pyramid,  are  found  by  Art.  58 ;  joining  these  points,  we  have 
efg  as  the  intersection  of  the  prism  with  the  face  abc  of  the 
pyramid.  The  three  elements  of  the  prism  do  not  leave  the  pyra- 
mid in  the  same  face ;  hence  find  the  points  w  aud  n  where  the 
elements  A  and  C  leave  the  face  ac d;  then  find  the  point  r  where 
the  element  B  would  leave  the  same  face  if  that  face  were  con- 
sidered of  indefinite  extent ;  joining  m,  w,  and  r,  we  have  the  inter- 
section of  the  prism  with  the  face  acd  when  extended  sufficiently. 
That  portion  of  the  triangle  mnts  which  actually  falls  on  the  lim- 
ited face  acd,  will  be  a  part  of  the  intersection  required;  the  ele- 
ment B  actually  leaves  the  pyramid  at  the  point  o  in  the  face  bed; 
joining  this  point  with  s  and  t,  the  intersection  is  completed. 

Fig.  72  shows  the  development  of  the  surface  of  the  pyramid 
with  the  intersections  traced  on  it. 

115.  Problem  40.  — To  construct  the  projections  of  a  right  hex- 
ugonal  prism  and  a  right  hexagonal  pyramid^  and  find  their  inter- 
section tvhen  they  occupy  a  certain  fixed  piosition  relative  to  each  other 
and  to  the  coordinate  planes.  —  Let  the  lowest  corner  of  the  prism 
rest  on  H  ;  the  long  diameter  of  the  base  of  prism  is  2^  inches ; 
its  height  is  3  inches ;  two  of  its  faces  are  in  a  plane  perpendicu- 
lar to  H ;  its  axis  makes  an  angle  of  60°  with  H  and  25°  with  V. 
The  long  diameter  of  the  base  of  the  pyramid  is  2  inches ;  its  height 
is  3  inches ;  its  axis  slopes  downward,  forward,  and  to  the  left,  mak- 
ing angle  of  45°  with  H,  and  whose  horizontal  projection  makes  an- 
gle of  15°  with  G  L.  A  point  in  the  axis  of  the  prism  2  inches  from 
the  base  is  ^  inch,  perpendicular  distance,  behind  a  point  in  the 
axis  of  the  pyramid  which  is  1  inch  from  the  base  of  the  pyramid. 

See  Fig.  73  for  construction.  If  the  axis  of  prism  makes  an  angle 
of  60°  with  H  and  25°  with  V,  the  plane  of  the  base  makes  an  angle 
of  30°  with  H  and  65°  with  V.  V  F  and  H  P,  the  traces  of  this 
plane,  are  found  by  Art.  69.  The  projections  of  the  prism  are 
then  found  by  Art.  71. 

Next  it  is  necessary  to  locate  the  axis  of  the  pyramid.  To  do 
this  we  use  the  converse  of  the  principles  in  Art.  65,  which 
explains  how  to  find  the  perpendicular  distance  between  two  lines 
when  their  projections  are  given  ;  here  the  projections  of  one  line, 


44  DESCRIPTIVE  GEOMETRY. 

the  direction  of  the  projections  of  the  other,  and  the  perpendicu- 
lar distance  between  them  are  given,  to  find  the  projections  of  the 
second,  d,  the  point  in  the  axis  of  the  prism,  2  inches  from  the 
base,  is  found  by  laying  off  on  the  revolved  position  of  the  axis 
rt,*  h!"  the  distance  a,*  dl'  equal  to  2  inches,  and  in  counter  revohi- 
tion  dj'  goes  to  d''.  Now,  through  this  point  d,  draw  the  line  A 
parallel  to  the  axis  of  the  p3'ramid  by  the  converse  of  the  princi- 
ples in  Art.  86.  A/  is  drawn  through  d\  making  angle  of  45" 
with  G  L.  A,''  must,  of  course,  be  parallel  to  G  L.  From  d"  lay 
off  on  A/  a  distance  d''  e,"  equal  to  2  inches,  and  f?"  n^  equal  to  1 
inch ;  in  counter  revolution  A,"*  moves  through  angle  of  15"  to 
A\  A/  moves  to  A^  ef  to  e",  ej"  to  g*,  w,"  to  n",  and  w  e  is  a 
line  through  d  equal  in  length  and  parallel  to  the  axis  of  the 
pyramid.  Through  the  line  A  and  the  axis  of  the  prism  pass  a 
plane.  V  Q  and  H  Q  are  the  traces  of  this  plane.  From  the 
point  d  draw  the  perpendicular  c?  o  of  indefinite  length ;  revolve 
this  indefinite  line  until  it  is  parallel  to  V,  and  lay  off  d''  o,"  equal 
to  i  inch  on  the  revolved  position  ;  in  counter  revolution  o,"  moves 
to  o",  and  o*  is  found  vertically  below  it  on  the  horizontal  projec- 
tion of  the  indefinite  perpendicular.  Through  the  point  o  draw 
the  \mefg  equal  and  parallel  to  ew,  and  this  line  will  be  the  axis 
of  the  pyramid  correctly  located  relative  to  the  axis  of  the  prism. 

Through  the  point  g  pass  the  plane  R  perpendicular  to  the  axis 
fg  of  the  pyramid.  This  will  contain  the  base  of  the  pyramid, 
and  the  projections  of  the  pyramid  can  then  be  found  the  same  as 
the  prism  by  Art.  71. 

The  intersection  of  the  two  bodies  is  obtained  by  finding  where 
each  edge  of  one  penetrates  the  faces  of  the  other,  if  at  all,  by 
Art.  58,  the  same  as  in  Art.  114. 

116.  Problem  41.  —  To  find  the  curve  of  intersection  of  two 
double  curved  surfaces  of  revolution  with  their  axes  intersecting.  — 
Let  the  two  surfaces  be  an  ellipsoid  and  a  sphere,  given  as  in 
Fig.  74,  their  axes  intersecting  at  o. 

Intersect  these  two  surfaces  by  a  series  of  auxiliary  spheres 
whose  centres  are  at  o;  each  sphere  will  cut  circles  out  of  the 
surfaces,  the  intersections  of  which  will  be  points  on  the  required 
curve.  The  sphere  A  cuts  out  of  the  ellipsoid  the  circle  C,  and 
out  of  the  sphere  the  circle  D,  their  intersections  e  and  e,  are  two 
points  on  the  curve.    Other  points  can  be  found  in  the  same  way. 


CHAPTER  VI. 

Miscellaneous  Problems. 

117.  Problem  42.  —  To  find  the  right  section  of  a  cylinder ^ 
making  a  certain  known  angle  with  the  horizon,  which  will  exactly 
mitre  ivlth  a  horizontal  cylinder  ivhose  axis  is  at  right  ayigles  to  it. 
—  Let  the  horizontal  cylinder  be  circular,  and  with  its  axis  per- 
pendicular to  V,  Fig.  75 ;  the  other  cylinder  makes  an  angle  ^ 
with  H,  and  is  parallel  to  V.  The  section  of  each  cylinder  on  the 
mitre  plane  Q  is  the  same. 

The  plane  P  will  cut  a  right  section  from  the  slanting  cylinder, 
whose  vertical  projection  will  be  in  the  line  V  P,  and  whose  hori- 
zontal projection  can  be  found  as  follows  j  The  plane  R  is  tangent 
to  both  cylinders  along  the  elements  A  and  B,  which  must  inter- 
sect at  the  point  a  in  the  plane  Q.  In  this  Way  A^  is  determined, 
and  consequently  6\  a  point  on  the  horizontal  projection  of  the 
curve.  In  the  same  Way  the  plane  T  gives  the  two  points  d  and 
d,.     Other  points  are  found  in  the  same  way. 

The  true  size  of  the  section  dh  d,  is  easily  found. 

118.  Problem  43.  — »I^o  find  the  right  section  of  the  raking 
moidding  ivhich  ivill  mitre  with  a  gutter,  the  pitch  of  the  roof  and 
the  section  of  the  gutter  heing  gi})en.-^'Y.\\\%  is  an  application  of  the 
principles  of  the  preceding  article. 

Let  the  section  of  the  gutter  be  given  as  in  Fig.  76,  and  the 
pitch  of  the  roof  equal  the  angle  ^. 

The  points  and  planes  have  been  lettered  the  same  as  in  Fig.  75, 
so  that  the  explanation  for  that  figure  is  equally  applicable  for 
this,  except  that  in  this  figure  the  true  size  of  the  section  is  shoWn 
as  revolved  into  a  plane  parallel  to  V  instead  of  into  H.  It  is 
not  necessary  that  the  back  part  of  the  moulding  be  made  to  fol- 
low the  inside  of  the  gutter,  as  it  Is  invisible ;  hence  it  is  made 
out  of  an  ordinary  plank  of  any  convenient  thickness,  and  the 
curve  exdh  {■&  aW  that  it  is  necessary  to  find. 

It  is  well  to  note  that,  when  the  plane  Q  makes  an  angle  of  45° 


46  I)ESCRrE»TtVE  GEOMETRY. 

with  V,  as  is  usual  in  such  work,  the  distances  d''  k,  x^  c,  etc.,  are 
equal  to  o  r,  s  f,  etc.,  respectively ;  hence  the  true  size  of  the  sec- 
tion can  be  obtained  directly  from  the  elevation  without  the  use 
of  the  plan,  as  follows:  Draw  any  number  of  lines,  as  V  R,  V  S, 
etc.,  making  angle  ^  with  the  horizon ;  draw  any  line,  as  V  P,  at 
right  angles  to  these  lines ;  from  the  points  of  intersection  c?",  re", 
etc.,  of  the  line  V  P  with  V  T,  V  X,  etc.,  lay  off  along  these  lines 
d"  d^,  x"-' x%  etc.,  equal  to  or.st^  etc.,  the  distance  from  the  points 
where  V  T,  V  X,  etc.,  intersect  the  curve  of  the  gutter  to  any  ver- 
tical line  as  C.  A  curve  drawn  through  the  points  d'\  x%  etc., 
will  be  the  curve  of  the  moulding. 

119.  Problem  W.~^Griven  the  right  section  of  the  doping  part 
qf  the  side  wall  to  a  flight  qf  steps,  to  find  the  right  section  of  the 
lower  end  of  this  wall  at  right  angles  to  the  given  section.  —  Let 
e  d^  x^  6,  Fig.  77,  be  half  the  given  section,  and  ^  the  angle  of  the' 
slope  of  the  wall. 

This  problem  is  simply  the  converse  of  the  preceding  one,  and 
needs  no  further  explanation. 

The  side  walls  of  the  Institute  steps  illustrate  this  problem. 

120.  Problem  45.  —  To  find  the  covering  of  an  eight-sided  dome 
as  given  in  Fig.  78. ' —  The  right  section  of  the  portion  o  ah  is  the 
curve  a"  t"  s",  etc.  This  right  section  unrolls  in  the  straight  line 
e''  t,  .*?,,  etc.,  e''  ^,,  t^  s,,  etc.,  being  equal  to  «"  (5",  f  s",  etc.,  respectively. 

The  right  section  of  the  portion  o  h  c  is  shown  at  5*  o',  and  it 
unrolls  in  the  straight  line  h*"  o,^^  b'^  o,/^  being  equal  in  length  to 
the  arc  h''  o'. 

The  right  section  of  the  portion  o  c  d  is,  the  ellipse  shown  at 
n''  o'\  which  unrolls  in  the  straight  line  w^  o,„*,  equal  in  length  to 
the  arc  w*  o". 

Since  most  of  the  remaining  problems  are  simply  applications 
of  principles,  already  familiar,  I  have  thought  it  best  to  give  the 
construction  without  the  description  in  some  cases ;  a  general 
description  without  the  construction  in  others ;  and  still  others 
without  either  construction  or  description,  in  order  that  the  stu- 
dent may  accustom  himself  to  apply  these  principles  to  the  solu- 
tion of  practical  problems  as  they  may  arise. 

121.  Problem  46.  —  Given  the  elevation  of  the  outline  of  a 
twelve-sided  vase,,  as  in  Fig^  79,  to  find  the  elevation  of  the  edges 


DESCRIPTIVE   GEOMETRY.  47 

A,  jB,  (7,  and  D,  thus  completing  the  elevation  of  the  object;  also 
the  development  of  one  of  the  sides. 

122.  Problem  47. — To  strike  the  pattern  of  the  sheet  metal 
used  in  making  a  bath-tub.  —  Figs.  80,  81,  and  82  show  the  con- 
struction for  three  different  styles  of  tubs. 

123.  Problem  48. — To  find  the  iyitersection  of  the  steam  dome 
and  boiler  of  a  locomotive  ;  to  develop  the  steam  dome  ;  and  also  to 
develop  the  taper  sheet.  —  Fig.  83  shows  the  two  elevations  of  a 
portion  of  a  locomotive,  and  the  intersection  of  the  steam  dome 
and  boiler. 

P^ig.  84  is  the  development  of  the  steam  dome,  and  Fig.  85  is 
the  development  of  the  taper  sheet  without  the  lap. 

124.  Problem  49.  — To  find  the  positioyi  of  a  guide  pidley,  and 
its  shaft,  to  take  a  belt  fi'om  one  pidley  to  another  whose  shafts 
are  at  right  angles  to  each  other.  —  Let  A  and  B,  Fig.  86,  be  the 
given  pulleys.  Assume  any  point  c  in  the  line  a  b  (as  the  guide 
pulley  may  be  placed  at  any  convenient  point  between  the  two 
given  pulleys)  ;  from  this  point  draw  the  lines  cm  and  c  e  tangent 
to  the  pulleys  B  and  A  respectively.  The  plane  P  of  these  two 
lines  c  m  and  c  e  will  be  the  plane  of  the  guide  pulley. 

The  ground  line  may  be  taken  in  any  convenient  position  per- 
pendicular to  the  given  shafts. 

To  find  the  projections  of  the  guide  pulley,  its  diameter  being 
known,  revolve  the  plane  P  into  the  horizontal  coordinate  plane ; 
cm  revolves  to  cf' fnf',  and  c  e  to  cf' e,'' ;  C}"  is  then  drawn  equal  to 
the  actual  size  of  the  guide  pulley.  In  counter  revolution  its 
centre  is  found  at  o"  o\  and  the  circle  C,**  is  found  at  C  C",  as 
explained  in  Art.  71. 

The  shaft  being  perpendicular  to  the  plane  of  the  pulley,  its 
projections  are  respectively  perpendicular  to  the  traces  V  P  and 
H  P,  and  they  must  pass  through  the  centre  o"  o'';  hence  through 
the  point  o  draw  the  indefinite  lines  o^n"  perpendicular  to  VP,  and 
o''  n''  perpendicular  to  H  P,  and  the  direction  of  the  shaft  will  be 
determined. 

To  show  a  definite  length  of  the  shaft,  revolve  the  line  rs 
until  it  is  parallel  to  V ;  then,  in  its  revolved  position,  lay  off 
o/  sj'  and  o,''  r,^  each  equal  to  half  the  length  of   the  shaft ;  in 


48  DESCRIPTIVE  GEOMETRY. 

counter  revolution  r,  goes  to   r,    s,  to  s,  and  rs  is  the  length 
required. 

125.  Problem  50. —  Given  the  longitudinal  section  and  plan^  or 
end  view,  of  a  connecting  rod  of  varying  cross  section,  to  complete  the 
drawing  of  the  elevation.  —  Let  the  section  and  plan  be  given  as  in 
Fig.  87.  The  lower  end  from  A  to  B  is  cylindrical,  from  C  to  D 
is  octagonal,  from  E  to  F  is  rectangular,  between  B  and  C  the 
cylinder  exj^ands  regularly  in  the  arc  of  a  circle  from  the  cylin- 
drical to  the  octagonal  section,  and  between  D  and  E  it  expands 
in  an  elliptical  curve  from  the  octagonal  to  the  rectangular 
section. 

126.  Problem  51. — Given  the  section  and  plan  of  a  portion  of 
a  rocker  arm,  as  in  Fig.  88,  to  complete  the  elevation.  — There  is  but 
one  curve  to  be  found,  and  that  is  simply  the  intersection  of  a 
cylinder  and  torus,  whose  axes  are  at  right  angles  with  each  other. 
The  cylinder  and  torus  are  shown  in  the  figure  by  dotted  lines. 

127.  Problem  52.  — To  find  the  development  of  the  conical  por- 
tion B  of  the  double  elbow  shown  in  Fig.  89,  the  right  sections  of  the 
portio7is  A  and  C  being  circular. 

128.  Problem  53. — To  find  the  section  on  AB  of  the  screw,  as 
given  in  Fig.  90 ;  also,  find  section  on  CI). 

129.  Problem  54.  —  To  construct  the  shadow  of  a  sphere  on  a 
horizontal  j^lati'^ }'  also,  find  its  line  of  shade.  —  Pass  a  plane  P, 
Fig.  91,  perpendicular  to  H,  through  the  ray  of  light  which  passes 
through  the  centre  of  the  sphere.  Revolve  this  jjlane  into  H. 
The  great  circle  which  it  cuts  from  the  sphere  revolves  to  the 
circle  of  which  o,^  is  the  centre ;  the  ray  of  light  through  the 
centre  revolves  to  of' a'';  cf  b''  and  c?/' o\  parallel  to  o/' a'',  are  the 
revolved  positions  of  the  two  rays  of  light  lying  in  the  plane  P, 
and  tangent  to  the  sphere ;  o''  and  6\  where  these  two  lines  inter- 
sect H  P,  are  the  two  extremities  of  the  long  diameter  of  the  ellipse 
which  is  the  shadow  of  the  sphere ;  the  short  diameter  of  this 
ellipse  e/'x''  %ill,  of  course,  be  equal  to  the  diameter  of  the  sphere. 

To  get  the  line  of  shade,  revolve  the  plane  P  back  to  its  orig- 
inal position  ;  the  point  cf'  goes  to  c^,  and  cZ,*  to  d'';  c*  and  d''  are 
two  points  of  the  ellipse  which  constitutes  the  line  of  shade. 


Fig.  2. 


*rf" 


4>r 


Fig.  13 


VQ 


Ic*       HP 


V      VP 


g!      Fig.49. 


Figr.47.      5! 


",' 

B"    i 

\     **- 

E"          a' 

c* 

A* 

Pig.56. 


'¥ig.58. 


t/ij  1 


Id' 


Fig.  62. 


m 


c 


-1^' 


Fig.66. 


Fig.65. 


Fig.7 1 


Pig.76. 


k      en!' 


i*'. 


-'«^X_i^ 


-^.v 


Fig.78. 


M ^ 


Fig.81. 


Fig.80. 


1_. 


Fig.84 


Fig.86. 


o 


ig.90. 


Fig.89. 


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